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FIRST  STEPS  IN  GEOMETRY 


BY 

G.  A.  WENTWORTH 

I' 

AND 

G.  A.  HILL 


lj>     \'»\     I      '\'     o          ' 

."'" 

A  i    5  ^^'M 

«  o'' 

;  ■ '  ' 

BOSTON,  U.S.A. 

GIKN 

&    COMPANY, 

PUBLISHERS 

Cbe  3ltl)cn3cum 

]J)re0fii 

1902 

Copyright,  1901,  by 
G.  A.  Wentworth  and  G.  A.  Hill 


ALL  RIGHTS  RESERVED 


A/)^' 


PKEFACE 

This  book  is  intended  to  be  an  introduction  to  elementary 
geometry.  It  aims  to  make  clear  by  illustrations,  definitions, 
and  exercises  the  exact  meaning  of  the  straight  line,  parallel 
lines,  axial  and  central  symmetry,  loci  of  points,  equal  figures, 
equivalent  figures,  similar  figures,  and  measurements  of  lines, 
surfaces,  and  solids. 

It  aims  also  to  make  the  learner  familiar  with  the  most  im- 
portant theorems,  and  to  teach  him  to  draw,  with  instruments 
and  free-hand,  accurate  figures  both  plane  and  solid. 

The  pupil  who  begins  the  study  of  any  of  the  common  text- 
books in  geometry  with  a  clear  knowledge  of  the  subject  matter 
of  this  book,  and  with  skill  in  drawing  geometric  figures  will 
have  a  great  advantage  over  the  pupil  who  begins  the  study 
without  such  knowledge  and  skill;  for  he  can  then  give  his 
whole  attention  to  the  processes  of  abstract  reasoning  employed 
in  the  demonstrations  of  propositions. 

We  acknowledge  our  indebtedness  to  Dr.  Francis  K.  Ball  of 
Phillips  Exeter  Academy,  who  has  read  all  the  proof  sheets,  and 
given  us  the  benefit  of  his  criticisms. 

A  box  of  drawing  instruments  can  be  had  of  the  publishers  at 
a  moderate  price.  The  student  should  also  have  a  small  T  square 
and  a  block  of  paper. 

G.  A.  WENTWORTH. 

G.  A.  HILL. 

April,  1901. 

lii 


M126SJ.3 


CONTENTS 

PAGE 

Geometric  Magnitudes 1 

Geometric  Magnitudes  and  Motion     .         .         ...  29 

Triangles  and  Quadrilaterals 56 

Circles  and  Regular  Polygons 72 

Areas ^3 

Similar  Figures 109 

The  Common  Geometric  Solids 137 

Index 153 

Answers 155 


iv 


FIEST  STEPS  IN  GEOMETEY 


CHAPTER   I 


GEOMETRIC   MAGNITUDES 


1.  Bodies.  This  block  of  wood  (Fig.  1)  is  called  a  body,  or 
a  solid,  because  it  occupies  space,  or  has  extension. 

The  block  is  also  called  a  cube,  on 
account  of  its  shape  or  form. 

The  cube  extends  in  three  princi- 
pal directions : 

from  A  to  B  (left  to  right), 
from  A  to  C  (front  to  back), 
from  A  to  D  (top  to  bottom). 

J^he  distances  from  A  to  B^  from  A 
to  C,  and  from  A  to  D  are  called  the 
dimensions  of  the  cube. 

One  of  the  dimensions  is  called  the  length,  another  the 
breadth,  and  the  third  the  thickness.  The  distance  from  top 
to  bottom  is  also  called  the  height,  or  the  depth. 

Every  body  has  three  dimensions.  If  the  dimensions 
differ  in  magnitude,  the  longest  one  is  usually  called  the 
length. 

Point  out  and  name  the  dimensions  of  this  book. 

Name  the  dimensions  of  this  room. 

1 


Fig.  1.  — a  Cube. 


2:  .':',/  ':'  J  FlR^l; ,aTEPS  in  geometry 

2.    A  cube  is  bounded  by  its  faces ;  the  faces  are  bounded 
by  the  edges ;  and  the  edges  are  bounded  by  the  corners. 

Fig.  2  represents  a  cube  drawn  in 
outline. 

How  many  faces  has  the  cube  ? 
How  many  edges  ? 
How  many  corners? 
The  faces  are  called  surfaces. 
Each  face  has  two  dimensions,  and 
only    two    dimensions,    length    and 
breadth,  or  length  and  height. 
Name  the  dimensions  of  the  top  face. 
Name  the  dimensions  of  one  of  the  side  faces. 
The  edges  of  the  cube  are  lines. 

A   line    has    one    dimension,    and    only    one    dimension, 
namely,  length. 

Each  edge  is  the  common  boundary  of  two  faces,  or  the 
intersection  of  two  faces. 

The  corners  of  the  cube  are  points. 

A  point  has  position,  but  no  extension  ;   it  has  no  length, 
breadth,  or  thickness. 

How  many  edges  meet  or  intersect  at  each  corner  ? 


Fig.  2.  — a  Cube  in  Outline. 


3.    A  point,  as   defined  in  geometry,  is  not  visible.     It 
is  represented   to    the   eye                             ^ 
on  paper  by  a  small  black 
dot,  as  shown  in  Fig.  3.  ■ 

A  line  is  represented  on     

paper  usually  by  a  narrow 


Fio.  3. 


mark,  or  by  a  series  of  short  dashes  placed  end  to  end,  as 
shown  in  Fig.  3. 


GEOMETRIC    MAGNITUDES  3 

4.  Straight  Lines.  Lines  are  of  two  kinds,  straight  and 
curved.  The  edges  of  a  cube  are  straight  lines.  A  string 
stretched  tight  represents  a  straight  line.  A  string  allowed 
to  hang  loose  takes  the  form  of  a 
curved  line,  or  curve. 

Fiff.  4  represents  a  straig-ht  line 

^  ^  ,  ^  .  Fig.  4. 

and   a   curved   line,    both    passing 

through  two  fixed  points,  C  and  E.  Suppose  that  these 
lines  are  made  to  revolve  about  the  fixed  points  C  and  JK. 
The  curved  line  will  take  different  positions,  but  the  straight 
line  will  keep  the  same  position. 

A  curved  line  changes  in  position  when  made  to  revolve 
about  two  of  its  points. 

A  straight  line  does  not  change  in  position  when  made  to 
revolve  about  two  of  its  points. 

Therefore,  it  follows  that 

Two  points  determine  a  straight  line  ;  that  is,  fix  its  position. 

It  also  follows  that 

Onli/  one  straight  line  can  be  drawn  from  one  point  to  another. 

A  point  C  (Fig.  5)  in  a  straight  line  divides  the  line  into 
two  parts,  CA  and  CB.  All  points  situated  on  the  part  CA 
are  said  to  have  the  same  direction  from  the  point  C;  and  all 
points  situated  on  the  part  CB  are  also  said  to  have  the  same 
direction  from  the  point  C. 

A^ + >B 

Fig.  5. 

The  parts  CA  and  CB  are  said  to  have  opposite  directions 
from  the  point   C. 

Every  straight  line  may  be  regarded  as  extending  indefi- 
nitely in  either  of  two  opposite  directions. 


FIRST   STEPS   IN   GEOMETRY 


Another  truth  characteristic  of  a  straight  line  in  distinction 
from  a  curved  line  is  illustrated  in  Fig.  6. 

— — • A  string  is  fastened  by  one  end  at 

B^  and  the  other  end  is  passed  over  a 

^      nail  at  A.    At  first  the  string  hangs 

I  /  in  the  form  of  a  curve.     If  we  pull 

V      I -/- 1 —  the  free  end  of  the  string,  the  part 

ill ^  \      between  A  and  B  gets  shorter  and 

"  shorter ;  when  it  is  as  short  as  pos- 

^^^"  ^"  sible,  it  is  a  straight  line.     Hence, 

A  straight  line  is  the  shortest  line  from  one  point  to  another. 

The  way  carpenters  apply  this  truth  for  the  purpose  of 

drawing  straight  lines  on  wood  by  means  of  a  chalk-line  is 

shown  in  Fig.  7. 


Pig.  7. 


The  distance  between  two  points  means  the  length  of  the 
straight  line  which  joins  them. 

For  drawing  straight  lines  on  paper  a  smooth  piece  of 
wood  with  a  straight  edge,  called  a  ruler,  is  used. 

1.  Mark  four  points  on  paper  and  connect  them  by  as  many  straight 
lines  as  possible. 

2.  Draw  a  straight  line  free-hand  and  test  it  with  the  ruler. 


GEOMETRIC    MAGNITUDES 


Fig.  8. 


5.    Plane  Surfaces.     Let  us  examine  the  faces  of  the  cube 
more  closely.    What  kind  of  surfaces  are  they  ?    The  answer 
is,    plane    surfaces,    or    planes.       They 
belong  to  the  same  class  of  surfaces 
as  the  floor  or  the  blackboard. 

On  the  other  hand,  the  surface  of  ^^^^ 
an  apple  (Fig.  8)  is  a  curved  surface. 
Many  lead-pencils  have  curved  sur- 
faces. Probably  you  can  find  other 
examples  of  plane  and  curved  sur- 
faces here  in  the  room.  Now  what 
is  the  difference  between  a  plane 
surface  and  a  curved  surface? 

Observe  the  blackboard.  You  can  choose  any  two  points 
on  its  surface  and  then  draw  on  the  surface  a  straight  line 
from  one  point  to  the  other.  Can  you  do  this  on  the  surface 
of  an  apple?  If  you  imagine  a  straight  line  joining  two 
points  of  the  surface  of  the  apple,  this  line  will  not  lie  on 

the  surface,  but  pass  directly 
through  the  apple  (Fig.  8). 

A  plane  surface  is  a  surface 
such  that,  if  a  straight  line  is 
drawn  between  any  two  points 
in  it,  this  line  will  be  wholly 
in  the  surface.     A  surface  that 
will  not  stand  this  test  is  called 
a  curved  surface. 
To  test  whether  a  surface  is  plane  (or  flat^  as  it  is  com- 
monly called),  hold  against  it  in  various  positions  the  straight 
edge  of  a  ruler  and  see  if  the  edge  has  unbroken  contact 
with  the  surface  (Fig.  9). 

Apply  this  test  to  the  face  of  a  wooden  cube. 


Fig.  9. 


6 


FIRST    STEPS   IN    GEOMETRY 


6.  Plane  Figures.  A  plane  surface,  bounded  by  one  or 
more  lines,  is  called  a  plane  figure. 

The  faces  of  a  cube  are  plane  figures,  and  each  face  is 
bounded  by  four  straight  lines. 

In  a  true  cube  all  these  lines  are  equal  in  length.  The 
edges  of  this  cube  look  equal,  but  a  closer  test  may  show 
slight  differences  in  length.     How  can  a  test  be  applied  ? 

First  Method.  Hold  a  strip  of  thin  paper  against  an  edge 
and  bend  the  strip  where  it  touches  the  ends  of  the  edge. 
Then  apply  the  strip  to  the  other  edges ;  and 
in  each  case  observe  whether  the  two  creases 
on  the  paper  coincide  with  the  ends  of  the  edge. 
Second  Method.  Apply  to  the  edges  the 
points  of  an  instrument  called  dividers  (Fig.  10). 
It  has  two  metal  legs  which  end  in  sharp  points, 
and  can  turn  about  a  pivot.  The  distance  be- 
tween the  points  can  be  regulated  at  pleasure 
by  turning  a  screw-head,  A.  Open  the  legs  till 
the  points  will  just  touch  the  ends  of  one  edge 
of  the  cube;  then  the  distance  between  the 
points  is  just  equal  to  the  length  of  this  edge. 
Apply  the  points  to  the  other  edges,  taking 
care  not  to  change  their  distance  apart,  and  thus  test  whether 
all  the  edges  are  exactly  of  the  same  length. 

If  you  find  small  variations  in  the  lengths  of  the  edges, 
either  you  have  not  been  careful  in  your  work,  or  the  cube 
has  not  been  properly  made  and  is  not  really  a  cube. 

1.  Draw  with  a  ruler  a  straight  line  of  any  convenient  length. 
Then  with  ruler  and  dividers  draw  a  straight  line  twice  as  long  ;  four 
times  as  long. 

2.  Draw  a  straight  line  which  appears  to  you  equal  to  one  edge. of  a 
cube.     Then  test  the  equality  by  means  of  dividers. 


Fig.  10. 


GEOMETRIC    MAGNITUDES 


7.  Let  us  now  make  on  paper  a  drawing  or  diagram  of  one 
face  of  the  cube.  Place  the  cube  on  the  paper  and  trace 
with  a  pencil  the  outline  of  that  face  which  lies  on  the 
paper  (see  Fig.  11). 

The  diagram  thus  made,  ABCD  (Fig.  12),  is  a  true  repre- 
sentation of  the  face  of  the  cube  both  in  size  and  in  shape. 
It  is  geometrically  equal  to  the  face. 

In  fact,  while  constructing  this  diagram  we  have  been 
giving  an  illustration  of  what  is  meant  by  geometric  equality. 


D 


Fig.  11. 


Fig.  12. 


Geometric  equality  does  not  mean  agreement  in  size  only, 
or  agreement  in  shape  only,  but  agreement  in  both  size  and 
shape.     Therefore, 

Two  plane  figures  are  equal  if  one  of  them  can  he  so  placed 
on  the  other  that  they  coincide  in  all  their  parts  and  form  a 
single  figure. 

Now,  if  our  cube  has  been  accurately  made,  we  can  place  the 
faces  one  after  another  on  the  diagram  ABCD  already  made, 
so  that  they  will  exactly  coincide  with  ABCD.     Therefore, 

1.  The  faces  of  a  cube  have  the  same  size. 

2.  The  faces  of  a  cube  have  the  same  shape. 

In  other  words,  the  faces  of  a  cube  are  equal  figures. 


8 


FIRST   STEPS   IN    GEOMETRY 


8.  Examine  one  face  of  a  cube  as  represented  by  the 
diagram  ABCD  (Fig.  13).  We  know  that  it  is  a  plane 
figure  bounded  by  four  equal  straight  lines.  But  this 
description  is  not  complete.  The  plane  figure  EFGH 
(Fig.  14)  is  also  bounded  by  four  equal  straight  lines,  yet 
in  shape  it  differs  greatly  from  a  face  of  a  cube. 

Take  a  narrow  strip  of  cardboard  to  represent  a  straight 
line  and  bend  it  so  as  to  make  a  square  frame  in  shape  like 
the  figure  ABCD.  You  can  easily  change  the  shape  into 
that  of  the  figure  EFGH.,  or  into  other  shapes. 


Fig.  13 


If  you  reflect  a  little,  you  will  see  that  these  differences  in 
shape  arise  from  the  different  ways  in  which  two  straight 
lines  can  meet  to  form  a  corner.  On  the  cube  they  meet  so 
as  to  form  what  are  called  square  corners,  or  right  angles. 
Right  angles  are  very  common.  You  can  easily  find  them 
on  a  sheet  of  paper  or  an  envelope,  or  on  the  floor,  sides,  and 
ceiling  of  a  room.  Very  likely  you  know  a  right  angle 
the  instant  you  see  it.  But  to  define  a  right  angle  is  not 
so  easy  a  matter.  We  cannot  hope,  however,  to  gain  a 
sound  knowledge  of  geometry  unless  we  know  the  exact 
meaning  of  every  word  which  we  use.  We  must  then  try 
to  define  a  right  angle. 

Let  us  first  explain  the  meaning  of  the  word  angle. 


GEOMETRIC    MAGNITUDES 


9 


Fig.  15. 


9.  Angles.  When  two  straight  lines  meet  they  form  an 
angle.  The  point  of  meeting  is  called  the  vertex  of  the  angle. 
The  lines  are  called  the  sides  of  the 
angle.  Thus,  the  straight  lines  AB^ 
AC  (Fig.  15),  meeting  at  A^  form  the 
angle  BAC^  or  the  angle  a.  Observe 
carefully  both  ways  of  naming  the 
angle. 

The  legs  of  the  dividers  (p.  6)  may  be  regarded  as  repre- 
senting the  sides  of  an  angle,  the  pivot  being  the  vertex. 
If  you  open  the  legs  a  little,  the  angle  which  they  form  is 
small;  open  the  legs  more  and  the  angle  becomes  larger. 
Evidently  angles  differ  in  magnitude. 

JEqual  angles  are  angles  which  can  he  so  placed  that  their 
vertices  coincide  and  their  respective  sides  coincide. 

This  definition  is  in  full  agreement  with  the  idea  of  geomet- 
ric equality  already  explained  on  page  7. 

When  two  straight  lines  intersect  (Fig.  16),  four  angles  are 
formed.     Name  them,  using  the  letters  in  Fig.  16. 

Name  two  angles  in  Fig.  16  unequal  in  magnitude. 

Are  any  of  the  angles  in  Fig.  16  equal  in  magnitude? 


H. 


o 


h- — I 

j  1 


Fig.  16. 


D 
Fig.  17. 


10.    Right  Angles.    Suppose  that  two  straight  lines  AB,  CD 
(Fig.  17)  intersect  each  other  so  that  the  four  angles  formed 


10  FIRST    STEPS   IN    GEOMETRY 

are  all  equal  in  magnitude ;  then  each  angle  is  called  a  right 
.angle,  and  the  two  lines  are  said  to  be  perpendicular  to  each  other. 
It  is  very  easy  to  make  four  right  angles,  using  only  your 
hands  and  a  sheet  of  paper,  EFGH  (Fig.  17). 

Fold  over  the  paper  so  that  the 
n, ^  ,        g^gg  j^jj  shall  coincide  with  the  edge 


E  D 

Fig.  17. 


FG.  Unfold  it  and  you  will  have 
made  a  crease  or  line,  CD.  Then 
fold  the  paper  so  that  the  edge  EF 
shall  coincide  with  the  edge  HG. 
Unfold  again,  and  you  have  the 
crease  AB.  The  two  creases  meet 
at  the  point  0  and  form  four  right 
angles. 

You  know  that  these  angles  are  all  equal,  because  in  the" 
act  of  folding  they  are  made  to  lie  upon  one  another  and 
coincide ;  and  this  is  the  test  of  geometric  equality. 

The  edges  of  a  cube  when  they  meet  form  right  angles. 

The  faces  of  a  cube,  therefore,  are  plane  figures  that  have 
four  right  angles  and  are  bounded  by  four  equal  straight  lines. 

The  faces  of  a  cube  are  called  squares;  and  a  cube  may 
be  defined  as  a  solid  bounded  by  six  equal  squares. 

Squares  may  differ  in  size  but  not  in  shape. 

Every  square  has  four  right  angles.,  and  is  hounded  hy  four 
equal  straight  lines,  called  the  sides  of  the  square. 

EXERCISES 

1.  How  many  right  angles  are  there  on  any  one  face  cff  a  cube? 

2.  How  many  right  angles  are  there  at  each  corner  ? 

3.  How  many  right  angles  are  there  on  a  cube  ? 

4.  How  many  edges  are  perpendicular  to  any  one  edge  ? 

5.  How  would  you  test  with  pencil  and  paper  whether  the  four 
angles  of  a  face  of  a  cube  are  all  equal? 


GEOMETRIC   MAGNITUDES 


11 


Fig.  18. 


B 


D 


11.    Parallel  Lines.     It  will  be  shown  in  Chapter  II  that 
two  straight   lines    perpendic- 
ular to  a  third   straight   line 
cannot  meet  however  far  they 
extend  (Fig.  18). 

Two  straight  lines  that  lie 
in  the  same  plane  and  cannot 
meet  however  far  they  extend  are  called  parallel  lines. 

Two  straight  lines  to  be  parallel  must  lie  in  the  same  plane. 

It  follows  that  the  opposite  sides  of  a  square  are  parallel, 
for  both  of  them  are  in  the  same  plane  and  perpendicular  to 
the  other  two  sides. 


Pig.  19.— a  Football  Ground. 


The  opposite  edges  of  a  ruler,  the  lines  on  ruled  paper,  and 
the  lines  on  a  football  ground  are  examples  of  parallel  lines. 


12  FIRST   STEPS   IN    GEOMETRY 

12.  Parallel  Lines  and  Planes.  A  straight  line  and  a 
plane  are  said  to  be  parallel  if  they  cannot  meet  however  far 
they  are  produced. 

Two  planes,  also,  are  said  to  be  parallel  if  they  cannot 
meet  however  far  produced. 

Find  examples  illustrating  these  definitions  by  examining 

1.  The  sm-face  of  a  cube. 

2.  The  surfaces  that  bound  the  room. 


13.    A  Straight  Line  Perpendicular  to  a  Plane.      Lay   a 

sheet  of  paper  on  the  table  and  draw  a  straight  line,  AOB 

(Fig.  20).  Hold  a  pencil  per- 
pendicular to  AOB^  with  the  point 
at  0.  Can  the  pencil  be  held  in 
more  than  one  position  perpendic- 
ular to  AOB  Sit  0? 

Draw  another  straight  line,  COD, 
and  hold  the  pencil  perpendicular 

p,j^j  20.  ^^  ^^^^  ^^^^^  ^^  ^^^  ^-^  ^^  ^-      ^^ 

more  than  one  position  of  the 
pencil  perpendicular  to  AB  and  CD  at  0  possible? 

The  pencil,  if  held  perpendicular  to  two  straight  lines 
drawn  through  0,  must  be  held  in  one  position  and  in  one  only. 
In  this  position  the  pencil  is  said  to  be  perpendicular  to  the 
plane  of  the  papei'. 

A  straight  line  is  'perpendicular  to  a  plane  when  it  is  per- 
pendicular to  any  two  straight  lines  that  can  he  drawn  in  the 
plane  through  the  point  of  its  intersection  with  the  plane. 

Give  examples  of  straight  lines  perpendicular  to  planes 
1.    On  a  cube.  2.    In  the  room.  3.    In  a  bookcase. 


GEOMETRIC    MAGNITUDES 


13 


14.  Two  Planes  Perpendicular  to  Each  Other.  Take  a 
piece  of  cardboard  and  bend  it  so  that  it  represents  two 
planes  intersecting  along  a  straight  line,  AB  (Fig.  21). 
From  any  point  0  in  AB  draw  perpendicular  to  AB  the 
straight  line  OC  in  one  plane  and  the  straight  line  01)  in 
the  other. 

When  the  planes  are  placed  as  in  Fig.  21,  the  angle  COD 
is  clearly  not  a  right  angle.     But  in  Fig.  22  we  see  two 


Fig.  21. 


Fig.  22. 


planes  so  placed  that  the  angle  COD  is  a  right  angle.     In 
this  position  the  planes  are  perpendicular  to  each  other. 

Two  intersecting  planes  are  per- 
pendicular, if  two  straight  lines, 
one  in  each  plane,  perpendicular 
to  the  line  of  intersection  of  the 
planes  at  the  same  point  form  a 
right  angle. 

A  carpenter  tests  whether  two         ,,_  „,     a  rp    o 

^  Fig.  23.  —  a  Try  Square. 

plane   surfaces  are    perpendicular 

by  applying  to  them  an  instrument  called  a  try  square  (Fig.  23). 


Give  examples  of  perpendicular  planes 
1.    On  a  cube.  2.   In  the  room. 


3.    In  a  bookcase. 


14 


FIRST   STEPS   IN   GEOMETRY 


15.  Vertical  and  Horizontal  Lines  and  Planes.  A  thread 
suspended  from  a  fixed  point  (Fig.  24)  and  supporting  a  piece 
of  lead  or  other  metal  is  called  a  plumb  line. 
A  plumb  line,  when  at  rest,  is  said  to  have 
a  vertical  direction. 

A  straight  line  or  a  plane,  parallel  to  a 
plumb  line,  is  called  a  vertical  line  or  plane. 

A  straight  line  or  a  plane,  perpendicular 
to  a  plumb  line,  is  said  to  be  horizontal. 

Straight  lines  and  planes  which  are  neither 
vertical  nor  horizontal  are  said  to  be  inclined. 

The  surface  of  still  water,  if  small  in 
extent,  is  nearly  horizontal ;  but  the  surface 
of  the  ocean  or  of  a  large  lake  is  curved, 
because  the  earth  is  round. 


Fig.  24. 


Fig.  25.  —  Appearance  of  Vessels  at  Sea. 


GEOMETRIC   MAGNITUDES 


15 


EXERCISES 


1.  Examine  the  floor,  walls,  and  ceiling  of  the  room,  and  point  out 
vertical  lines  and  planes.     Point  out  also  horizontal  lines  and  planes. 

2.  What  kind  of  a  plane  is  the  football  ground  in  Fig.  19,  p.  11  ? 
How  would  you  describe  the  five-yard  lines  ?  the  goal  posts  ? 

3.  How  would  you  describe  the  surface  of  a  pond  when  the  water  is 
at  rest  (Fig.  26)  ?  AVhat  kind  of  line  does  a  pole  floating  on  the  water 
represent?  Will  it  make  any  difference  if  the  wind  blows  the  pole 
into  a  new  position  ? 

4.  What  is  there  in  the  appearance  of  the  vessels  in  Fig.  25  which 
shows  that  the  surface  of 

the  ocean  is  curved? 

5.  Point  out  vertical 
lines,  vertical  planes,  hori- 
zontal lines,  horizontal 
planes,  inclined  lines,  and 
inclined  planes  on  your 
desk. 

6.  What  kind  of  plane 
surface  is  represented  by 
a  door  ?  Does  it  make  any 
difference  whether  the  door 
be  open  or  shut? 

7.  Place  a  cube  upon 
the  table.  In  this  position, 
how  many  of  its  edges  are 
vertical?  How  many  are 
horizontal  ?  How  many  of 
its  faces  are  vertical  ?  How 
many  horizontal  ? 

8.  Hold  a  cube  so  that  four  edges  shall  be  horizontal  and  all  the 
other  edges  inclined  to  the  horizon. 

9.  Hold  a  cube  so  that  all  its  edges  shall  be  inclined  to  the  horizon. 

10.  Describe  the  positions  of  the  hands  of  a  clock  at  8  a.m. 

11.  Mention  the  times  of  day  when  the  hour  4iand  of  a  clock  is 
horizontal.      What  is  the  position  of  the  minute  hand  at  these  times  ? 


Fig.  2G.  —  Surface  of  Still  Water. 


16 


FIRST   STEPS   IN   GEOMETRY 


^P 

^^^^^3 


16.  Plumb  Rule.  The  walls  of  buildings  should  be  ver- 
tical ;  for  if  this  is  not  the  case,  they  are  in  danger  of  falling. 

Masons,  when  constructing  a 
brick  wall,  test  whether  it  is  ver- 
tical by  holding  against  it  a  plumb 
rule  (Fig.  27).  The  edges  of  tlie 
plumb  rule  are  made  parallel,  and 
a  straight  line,  called  the  test  line, 
is  drawn  on  the  wood  exactly  mid- 
way between  the  edges.  A  plumb 
line  is  suspended  at  A,  . 

The  workman  holds  one  edge  of 
the  plumb  rule  against  the  wall  to 
be  tested ;  if  the  wall  is  vertical, 
the  plumb  line  will  coincide  with  the  test  line  on  the  wood. 
To  complete  the  test,  the  experiment  is  repeated  by  holding 
the  opposite  edge  of  the  rule  against  the  wall. 

17.  Spirit  Level.  To  test  whether  a  surface  is  horizontal, 
a  spirit  level  is  the  instrument  in  common  use.  It  consists 
of    a    wooden    bar  


Fig.  27.  — Testing  a  Brick  Wall. 


^^^ 


Fig.  28.  — Testing  a  Horizontal  Plane. 


having  a  plane  sur- 
face   on    its    lower 
side,  and  on  its  up- 
per side  a  glass  tube     : 
slightly  curved  and 
nearly  full  of  alco- 
hol.    The  space  not  filled  with  alcohol  appears  as  a  bubble 
of  air.     When  the  bottom   of  the  level  is   horizontal,   the 
bubble  will  stand  exactly  at  the  middle  of  the  tube. 

Take  a  level  and  test  whether  the  surface  of  a  table  is  horizontal. 
Is  it  sufficient  to  apply  the  level  in  only  one  position  to  the  sm-face  of 
the  table  ? 


GEOMETRIC   MAGNITUDES 


17 


18.  Development  of  the  Surface  of  a  Cube.  Let  us  make 
a  diagram  of  the  entire  surface  of  a  cube,  as  it  would  appear 
if  spread  out  upon  a  plane  surface. 

Place  the  cube  on  a  sheet  of  paper  and  trace  the  outline  A  BCD 
(Fig.  29)  of  the  face  which  lies  on  the  paper.  Turn  the  cube  over  the 
edge  CD  till  another  face  lies  on  the  paper.  Trace  its  outline  CDEF. 
Similarly  trace  the  outlines  of  two  more  faces,  EFGH  and  GHIK. 
Then  place  the  cube  as  it  was  when  standing  upon  CDEF  and  trace  the 
two  remaining  faces  CFLM  and  DENO  by  turning  the  cube  over  the 
edges  CF  and  DE. 

M 


D 


K 


E 


H 


TV- 
Fig.  29.— Development  of  the  Surface  of  a  Cube. 

The  entire  diagram,  consisting  of  six  squares  Tying  in  one 
plane,  is  called  the  development  of  the  surface  of  the  cube. 

If  the  diagram  is  made  on  thin  cardboard,  and  if  the 
squares  are  then  properly  bent  about  their  sides,  and  their 
loose  edges  fastened  together,  a  model  of  a  cube  will  be  the 
result. 

Instead,  however,  of  constructing  the  development  of  the 
surface  as  just  explained,  it  is  better  to  make  use  of  a  ruler, 
dividers,  and  an  instrument  called  a  triangle.  The  triangle  is 
shown  in  the  diagram  on  the  next  page. 


18 


.    FIRST   STEPS   IN   GEOMETRY 


19.    Ruler  and  Triangle.     A  ruler  and  a  triangle  (Fig.  30) 
are  used  together  for  the  purpose  of  drawing  rapidly  straight 

lines  parallel  or  perpendicular  to  one 

another.       Both    instruments    have 

straight   edges,    and   the    edges    CD 

and  ED  of   the  triangle  should  be 

exactly  perpendicular  to  each  other. 

Place  the  ruler  and  the  triangle  as 

seen   in   Fig.  30,  slide  the    triangle 

along  AB^  and  in  different  positions 

draw  straight  lines  along  the  edge  CD  of  the  triangle.     These 

lines  will  be  perpendicular  to  AB  and  parallel  to  one  another. 

We  proceed  to  explain  the  method  of  constructing  a  model 

of  a  cube  with  the  aid  of  ruler,  triangle,  and  dividers. 


Fig.  30.  — Ruler  and  Triangle. 


Take  a  piece  of  cardboard.  Draw  a 
straight  line,  AE  (Fig.  31).  Lay  ofE  with 
dividers  the  lengths  AB,  BC,  CD,  DE, 
each  equal  to  an  edge  of  the  cube.  Com- 
plete the  figure,  using  ruler  and  triangle 
for  drawing  the  lines,  and  the  dividers 
for  measuring  lengths.  Draw  small  laps 
on  seven  edges,  as  seen  in  Fig.  31.  The 
laps  are  use«t  to  fasten  together  the  edges 
after  folding. 

Cut  out  the  figure,  and  fold  on  the  lines 
that  are  represented  as  dotted  lines  in  the 
figure.  Apply  mucilage  or  glue  sparingly 
to  the  laps,  and  fold  so  that  the  laps 
shall  come  on  the  inside  of  the  faces. 


\B 


y- 


E' 


Fig.  31. 


EXERCISES 

1.  How  many  cubes,  all  of  the  same  size  as  the  model,  must  be  put 
together  to  make  a  cube  of  larger  size  than  the  model  ? 

2.  How  do  the  lengths  of  the  edges  of  the  model  and  new  cube  compare? 


GEOMETRIC    MAGNITUDES 


19 


20.    Units  of  Length.     You  are  now  prepared  to  under- 
stand the  great  convenience  of  employing  units  of  length. 
The  standard  units  of  length  in  common  use  are  : 
the  inch  (in.),  the  foot  (ft.),  the  yard  (yd.),  and  the  mile. 
12  in.  =  1  ft. ;    3  ft.  =  1  yd. ;    5280  ft.  =  1  mile. 

The  inch  is  divided  into  halves,  quarters,  eighths,  and 
sixteenths. 

A  line  is  measured  by  finding  the  number  of  units  of  lengtli 
it  contains.  The  number  of  units  with  the  name  of  the  unit 
is  called  the  length  of  the  line. 

Short  lines  are  measured  by  a  graduated  ruler. 

A  part  of  a  graduated  ruler  is  shown  in  Fig.  32. 


IMlllllllniMlllMllllMlllllllllllMllllllllllll 

Fig.  32.  — Ruler  Graduated  to  Sixteenths  of  an  Inch. 

As  exercises  merely  in  estimating  lengths  by  the  eye,  give 
the  length  of  your  teacher's  desk  ;  the  length  of  the  room  ; 
the  width  of  tlie  room  ;   the  length  of  the  blackboard. 

Pacing  is  a  rough  but  quick  way  of  measuring  distances. 
The  two  following  exercises  should  be  done  by  you  before 
the  next  recitation,  and  the  results  reported  at  that  time. 


EXERCISES 

1.  Find  the  average  length  of  your  pace.  To  do  this  take  10  steps 
as  you  naturally  walk,  measure  the  distance  in  feet,  and  divide  this  dis- 
tance hy  10.  Repeat,  and  if  the  results  differ,  find  the  mean  result 
by  adding  the  two  results  and  taking  half  the  sum. 

2.  Find,  by  pacing,  the  distance  from  your  home  to  your  school, 


20 


FIRST   STEPS   m   GEOMETRY 


Some  exercises  in  measuring  lines  will  now  be  given.  In 
each  case,  begin  by  trying  to  estimate  the  length  to  be  meas- 
ured as  well  as  you  can,  then  measure  the  line  correct  to  a 


Line  to  be  Measured. 

Estimated 
Length. 

Measured 
Length. 

Difference. 

Edge  of  Cube  .     . 
Length  of  Page     . 
Etc 

sixteenth  of  an  inch  with  a  graduated  ruler,  and  make  a 
record  of  your  work  in  the  form  of  a  table,  as  shown  above. 


EXERCISES 

1.  Measure  the  edge  of  the  cube  given  you. 

2.  Measure  the  length  of  this  page. 

3.  Measure  the  width  of  your  desk. 


Fig.  33. 

4.  Measure  the  length  of  a  printed  line  on  the  page. 

5.  Measure  the  line  AB  in  Fig.  33. 

6.  Measure  the  line  BCin  Fig.  33. 

7.  Measure  the  line  ^  C  in  Fig.  33. 

8.  Measure  the  line  DE  in  Fig.  33. 

9.  Measure  the  line  EF  in  Fig.  33. 


GEOMETRIC    MAGNITUDES 


21 


21.  The  Metric  System  of  Units.  The  metric  units  of 
length  are  the  millimeter  (mm),  the  centimeter  (cm),  the  deci- 
meter (dm),  the  meter  (m),  and  the  kilometer  (km). 

They  form  a  part  of  the  Metric  System  of  units,  which  is 
throughout  a  decimal  system. 

10  mm  =  1  cm,        10  dm  =  1  m, 
10  cm  =1  dm,     1000  m  =  1  km. 
A  ruler  graduated  to  millimeters  is  shown  in  Fig.  34. 


12           3           4 

5           6           7 

8 

Fig.  34.— Ruler  graduated  to  Millimeters. 

A  length  expressed  in  any  metric  unit  is  reduced  to  the 
next  smaller  unit  by  multiplying  the  number  by  10,  and  to  the 
next  larger  unit  by  dividing  the  number  by  10  ;  the  first 
reduction  is  made  by  moving  the  decimal  point  one  place  to 
the  right,  the  second  by  moving  it  one  place  to  the  left. 

Thus  3.25  m      =  32.5  dm    =  325  cm     =  3250  mm. 

And  6452  mm  =  645.2  cm  =  64.52  dm  =  6.452  m. 

In  recording  a  measurement  in  metric  units,  use  only  one 
unit,  and  employ  the  decimal  point  when  necessary.  For 
example,  if  the  length  of  an  edge  of  a  cube  is  9  centimeters 
and  6  millimeters,  record  it  either  as  9.6  cm,  or  96  mm. 


EXERCISES 

1.  How  many  millimeters  are  there  in  1  dm?   in  4  dm?   in  1  m? 

2.  How  many  centimeters  are  there  in  1  m?   in  6  m?   in  35  mm? 

3.  How  many  decimeters  are  there  in  7  m?    in  20  cm?   in  200  mm? 

4.  How  many  meters  are  there  in  50  dm?   in  225  cm?   in  800  mm? 
Perform  again  the  first  seven  exercises  on  page  20,  using  metric  units, 


22 


FIRST   STEPS   m   GEOMETRY 


22.  The  Square  Prism.  The  solid  in  Fig.  35  is  called  a 
square  prism. 

How  many  faces  has  it  ?  How  many  edges  has  it  ?*  How 
many  corners  has  it? 

The  upper  and  lower  faces  are  called  the  bases. 

The  other  faces  are  called  the  lateral  faces. 

The  edges  connecting  the  bases  are  called  the  lateral  edges. 


/ 

/ 

Fig.  35.  —  a  Square  Prism.  Fig.  3G.  —  Development  of  a  Square  Prism. 

The  lateral  edges  all  have  the  same  length.  The  edges  of 
the  bases  have  the  same  length.  But  a  lateral  edge  and  a 
base  edge  do  not  have  the  same  length,  hence  the  lateral 
faces  are  not  squares. 

The  faces  have  four  right  angles,  but  they  do  not  have  four 
equal  sides.     They  are  called  rectangles. 

Make  a  model  of  a  square  prism  precisely  as  you  made  a 
model  of  a  cube  (p.  18).  Make  each  lateral  edge  7  cm  long 
and  each  base  edge  4  cm  long. 

If  you  cut  a  square  prism  into  two  equal  parts  by  a  plane  surface 
parallel  to  the  bases,  what  kind  of  solids  will  the  parts  be? 


GEOMETRIC    MAGNITUDES 


23 


23.  The  Rectangular  Prism, 
called  a  rectangular  prism.  It 
parallelopipedon . 

In  what  respects  is  it  like  a 
square  prism  ?  In  what  respects 
does  it  differ  from  •  a  square 
prism  ? 

Are  all  the  angles  on  the 
prism  right  angles? 


IS 


The    solid   in    Fig.   37    is 
also    called    a   rectangular 


Fig.  37.  —  A  Rectangular  Prism. 


Are  all  the  faces  of  the  prism 
rectangles  ? 

Make  a  model  of  a  rectangular 
prism. 

Make  the  length  12  cm  (4i 
in.),  the  breadth  8  cm  (3  in.),  and 
the  height  4  cm  (li  in.). 

The  development  of  the  sur- 
face is  shown  in  Fig.  38. 

A  prism,  either  square  or 
rectangular,  is  a  very  common 
form  in  which  things  are  made. 
A  shelf  of  a  bookcase  is  a  rectan- 
gular prism.  A  sheet  of  paper, 
when  its  thickness  is  considered,  is  a  rectangular  prism. 
Give  other  examples  of  prisms. 

EXERCISES 

1.  Point  out  on  your  model  three  pairs  of  parallel  faces. 

2.  Point  out  on  your  model  three  groups  of  parallel  edges. 

3.  If  the  prism  is  cut  in  two  by  a  plane  through  AB  and  BC  which 
are  parallel  to  the  base,  and  also  by  a  plane  through  DE  perpendicular 
to  the  plane  ABC,  and  EF  perpendicular  to  ED,  what  kind  of  solids 
are  the  parts? 


( 

/ 

\ 

\ 

\                                            / 

Fig.  38. 


24 


FIRST   STEPS   IN   GEOMETRY 


24.  Triangular  Prisms.  If  a  square  prism  (Fig.  39)  is  cut 
in  two  along  the  lateral  edges  AB  and  CD,  it  will  be  divided 
into  two  parts,  equal  in  size,  called  triangular  prisms. 


Fig.  39.  —  Square  Prism.  Fig.  40.  —  Triangular  Prisms. 

Triangular  prisms  are  represented  in  Fig.  40. 

The  bases  of  a  triangular  prism  are  plane  figures  bounded 
by  three  straight  lines. 

Plane  figures  which  are  bounded  by  three  straight  lines 
are  called  triangles. 

A  triangle  having  a  right  angle  is  called  a  right  triangle. 


EXERCISES 

1.  Point  out  on  the  square  prism  parallel  planes;  perpendicular 
planes ;  parallel  lines ;  perpendicular  lines ;  lines  parallel  to  planes ; 
lines  perpendicular  to  planes. 

2.  Hold  the  prism  so  that  one  of  the  lateral  faces  shall  be  hori- 
zontal. Then  describe  the  position  of  all  the  other  faces,  and  also  of 
all  the  edges. 

3.  Hold  the  prism  so  that  all  of  its  edges,  and  also  all  of  its  faces, 
shall  be  inclined  to  the  horizon. 

4.  Are  the  bases  of  a  triangular  prism  equal  or  unequal  figures? 

5.  Describe  how  you  would  test  their  equality. 

6.  What  kind  of  triangles  are  they  ? 

7.  How  many  lateral  faces  has  a  triangular  prism  ? 

8.  What  kind  of  figures  are  the  lateral  faces  ? 

9.  Compare  the  lateral  faces  in  respect  to  magnitude. 


GEOMETRIC   MAGNITUDES 


25 


To  make  a  model  of  a  triangular  prism,  construct  the 
development  of  the  surface  as  follows : 

Draw  a  straight  line,  and  on  this 
line  take 

^^  =  4  cm  (11  in.), 
BC  =  8  cm  (3  in.), 
CD  =  4:  cm  (11  in.). 

Draw  through  B  and  C  lines 
perpendicular  to  AD. 

Take  BE,  BF,  CG,  CH,  each  4  cm 
(H  in.). 

Draw  the  straight  lines  A  F,  DH, 
EG. 

Extend  BF  and  CH. 

Take  FK  equal  to  FA,  and  HL 
equal  to  HD. 

Draw  KL. 

Draw  laps  as  shown  in  the  figure. 

Cut  out  the  figure  and  fold  on  the  required  lines. 

Fasten  by  means  of  the  laps. 

Prisms  may  differ  in  the  number  of  edges  that  bound  their 
bases ;  but  all  prisms  agree  in  having  for  bases  two  parallel 
and  equal  plane  figures,  and  in  having  for  lateral  faces  plane 
figures  bounded  by  two  pairs  of  parallel  lines. 


/ 

A 

/ 

\ 

/' 

\ 

K 
L 

F 
H 

B 

C 

E 
G 

\ 

/ 

\ 

\ 

/ 

D 

Fig.  41. 


EXERCISES 

1.  Point  out  parallel  faces  on  your  model ;  parallel  edges. 

2.  Point  out  perpendicular  faces ;  perpendicular  edges. 

3.  How  many  right  angles  in  all  are  formed  by  the  edges  ? 

4.  How  many  angles  are  there  that  are  not  right  angles  ? 

5.  Hold  the  model  so  that  six  edges  may  be  horizontal. 

6.  Hold  the  model  so  that  five  edges  may  be  horizontal. 

7.  Hold  the  model  so  that  three  edges  may  be  horizontal. 

8.  Hold  the  model  so  that  two  edges  may  be  horizontal. 


26  FIRST   STEPS   IN   GEOMETRY 

25.  Geometric  Bodies.  In  the  study  of  geometry  a  sharp 
distinction  is  drawn  between  the  form  of  a  body  and  the 
matter  of  which  it  is  composed.  The  form  may  be  that  of 
a  cube,  a  prism,  a  sphere,  etc. ;  the  matter  may  be  wood,  or 
iron,  or  glass,  or  some  other  substance. 

More  is  implied  in  this  distinction  than  appears  at  first 
sight.  Take  two  cubes,  one  made  of  wood,  the  other  of 
iron.  Not  only  do  they  differ  in  their  properties  because 
they  are  composed  of  different  materials,  but  it  is  also  true 
that  neither  of  them,  however  carefully  made,  is  a  perfect 
cube.  Human  skill  is  incapable  of  making  their  faces  per- 
fect planes,  their  edges  perfect  straight  lines,  their  corners 
perfect  points,  or  their  angles  perfect  right  angles. 

We  can,  however,  easily  imagine  such  a  thing  as  a  perfect 
cube ;  we  have  only  to  form  a  mental  picture  of  a  portion  of 
space  bounded  by  six  perfect  squares,  absolutely  equal  one 
to  another. 

This  geometric  cube,  as  we  may  call  it,  exists  indeed  only 
in  the  mind ;  but  it  is  a  far  simpler  object  of  study  than  any 
material  cube  ;  for  it  has  no  other  properties  than  those  which 
are  connected  with  its  form.  And  these  properties  when 
discovered  can  be  stated  as  absolute  truths  because  the  cube 
which  we  study  is  a  perfect  cube. 

Geometry  is  a  science  in  which  we  study  bodies  with 
respect  to  form  and  position  only.  In  order  to  make  the 
study  possible,  we  substitute  in  thought  for  material  bodies 
ideal  forms  enclosing  space,  known  as  geometric  bodies.  When 
in  this  way  a  science  of  geometry  has  been  constructed,  the 
truths  discovered  can  be  applied  to  useful  purposes  by 
substituting  for  actual  material  bodies  the  ideal  forms  that 
the  actual  bodies  most  closely  resemble. 


REVIEW   OF   CHAPTER   I 


27 


REVIEW  EXERCISES 

1.  Write  the  best  definitions  you  can  of  the  terms  below.  Try  to 
make  your  definitions  perfectly  clear  and  to  use  as  few  words  as 
possible. 


Body 
Surface 
Line 
Point 

Straight  Line 
Plane  Surface 
Plane  Figure 


Angle 

Right  Angle 
Square 
Rectangle 
Triangle 
Right  Triangh 
Parallel  Lines 


Perpendicular  Lines 
Parallel  Planes 
Perpendicular  Planes 
Horizontal  Line 
Vertical  Line 
Horizontal  Plane 
Vertical  Plane 


2.  When  is  a  straight  line  parallel  to  a  plane  ? 

3.  When  is  a  straight  line  perpendicular  to  a  plane  ? 

4.  Can  two  horizontal  planes  intersect  each  other? 

5.  Can  two  vertical  planes  intersect  each  other? 

6.  Can  a  horizontal  plane  intersect  a  vertical  plane  ? 

7.  What  kind  of  a  line  is  the  intersection  of  two  vertical  planes? 

8.  How  many  vertical  lines  can  be  drawn  through  a  point  ?     How 
many  horizontal  lines  ?     How  many  inclined  lines  ? 

9.  How  many  horizontal  lines  can  be  drawn  in  a  vertical  plane? 
How  many  in  a  horizontal  plane  ? 

10.  How  many  vertical  lines  can  be  drawn  in  a  horizontal  plane? 
How  many  in  a  vertical  plane? 

11.  Draw  three  straight  lines  so  that  they  intersect  one  another 
in  three  i)oints;  in  two  points;  in  one  point. 

12.  Through  a  given  point  how  many  straight  lines  parallel  to  a 
given  straight  line  can  be  drawn? 

13.  Draw  a  straight  line.  Then  place  three  points  so  that  you  can 
draw  through  each  one  of  them  a  line  parallel  to  the  straight  line 
already  drawn. 

14.  How  many  pairs  of  parallel  Wnen  can  be  drawn  througli  two 
given  points? 

15.  Through  a  given  point  how  many  straight  lines  perpendicular 
to  a  given  straight  line  can  be  drawn  ? 


28  FIRST    STEPS   IN   GEOMETRY 

16.  Draw  a  straight  line,  and  then  place  three  points  so  that  only- 
one  straight  line  can  be  drawn  perpendicular  to  the  line  already  drawn 
and  passing  through  all  three  points. 

17.  How  many  straight  lines  can  be  drawn  from  a  given  point  to 
a  given  straight  line  ?  Describe  the  position  of  the  shortest  of  these 
lines. 

18.  How  many  right  angles  are  formed  when  two  perpendicular 
lines  intersect  each  other? 

19.  Mention  the  times  of  day  when  the  hands  of  a  clock  form  a 
right  angle. 

20.  Is  the  angle  formed  by  the  han^s  of  a  clock  greater  or  less  than 
a  right  angle  at  10  a.m.?  at  4  p.m.? 

21.  How  many  times  during  24  hours  do  the  hands  of  a  clock  make 
the  same  angle  as  at  10  a.m.  ? 

22.  What  kind  of  a  straight  line  is  a  line  running  north  and  south  ? 
a  line  running  east  and  west  ? 

23.  Draw  a  straight  line,  AB,  4:  in.  long.  At  A  erect  a  perpendicu- 
lar, ^C  =  1^  in.  At  i5  erect  a  perpendicular,  BD  =  4^  in.  Join  CD; 
measure  its  length  ;   write  the  value  of  the  length  by  its  side. 

Work  this  exercise  first  free-hand,  and  then  with  instruments. 

24.  Draw  a  straight  line,  ^^  =  5  in.  Erect  at  ^  a  perpendicular, 
BC  =  S  in.  Draw  through  C  a  straight  line,  CD  =  1  in.,  and  parallel 
to  AB.     Join  AD  and  measure  its  length. 

Work  this  exercise  first  free-hand,  and  then  with  instruments. 


CHAPTER   II 

GEOMETRIC   MAGNITUDES   AND   MOTION 

26.  Path  of  a  Moving  Point.  When  we  move  the  point 
of  a  pencil  along  a  sheet  of  paper  the  point  of  the  pencil 
leaves  a  black  trace  which  we  call  a  line.  It  is  not  a  line, 
however,  but  a  long,  narrow  body. 

Imagine  the  point  of  the  pencil  to  be  a  point  as  understood 
in  geometry ;  then  the  trace  will  represent  a  geometric  line, 
having  length  as  its  sole  dimension. 

We  may  regard  any  line  as  made  or  generated  by  a  moving 
point,  and  define  a  line  as  the 
path  of  a  moving  point.     This 
way  of   regarding   a   line    is 
often  very  useful. 

The  generation  of  a  line 
by  a  moving  point  is  forcibly 
illustrated  when  a  small  lumi- 
nous body,  as,  for  example, 
the  red-hot  end  of  a  poker,  is 
put  in  rapid  motion  in  a  dark 
room.  The  end  of  the  poker 
appears  to  be  changed  to  a 
line  of  fire.     The  following  experiment  is  still  more  striking. 

Mount  a  Chinese  incense  stick  on  an  axis  (Fig.  42),  so 
that  its  ends  are  unequally  distant  from  the  axis.  Light 
the  ends  and  set  the  stick  in  rapid  rotation.  Instantly  in 
place  of  the  luminous  points  two  bright  rings  are  seen. 

29 


30  FIRST   STEPS   IN   GEOMETRY 

This  effect  is  explained  by  the  fact  that. the  impression 
of  light  produced  on  the  retina  of  the  eye  lasts  long  enough 
for  the  luminous  point  to  make  one  revolution,  and  thus 
renew  the  impression  before  it  has  time  to  die  out. 

27.  Path  of  a  Moving  Line.  A  moving  line,  in  general, 
generates  a  surface. 

Mount  a  glass  tube  upon  an  axis  so  that  the  tube  is  per- 
pendicular to  the    axis,   and   the    axis    passes  through   the 

middle  point  of  the  tube  (Fig. 
43).  When  at  rest,  the  tube 
represents  a  straight  line.  But 
when  set  in  rapid  rotation,  the 
tube  will  appear  to  be  trans- 
formed into  a  well-known 
plane  figure  called  a  circle. 
This  plane  figure  is  the  path 
traversed  by  the  tube  during 
each  revolution  about  the  axis. 
Reduce  in  thought  the  tube  to 
^^^'  *^'  a  geometric  line,  and  its  path 

will  be  reduced  to  a  geometric  magnitude  of  two  dimensions, 
that  is,  to  a  surface. 

28.  Path  of  a  Moving  Surface.  A  moving  surface,  in  gen- 
eral, generates  a  solid.     If  the  surface 

ABCD  (Fig.  44)  is  moved  to  the  right  ^  ^ 

to  the  position  EFGH^   the  surface 

ABCD  will  generate   the  solid  AG. 

The    lines    AB,    BC,    CD,    and    DA        [/ 

will  generate  the  surfaces  At\  BG,  f     44 

CH,  and  DE,  and  the  points  A,  B, 

(7,  and  D  will  generate  the  lines  AE,  BE,   CG,  and  DIf. 


E/ 


GEOMETRIC   MAGNITUDES   AND   MOTION  31 

29.  The  Circle.  Suppose  that  a  straight  line  OA  (Fig.  45) 
revolves  about  the  point  0,  keeping  always  in  the  same 
plane,  till  it  returns  to  its  first  posi- 
tion. It  will  generate,  or  describe,  a 
plane  figure,  called  a  circle.  The  point 
A  will  describe  a  curved  line,  which  is 
the  boundary  of  the  circle,  and  is  called 
the  circumference  of  the  circle. 

The  point  0  about  which  the  line  re- 
volves is  called  the  centre  of  the  circle. 

The  circumference  of  a  circle  is  the  path  of  a  point  moving 
in  a  plane  at  a  given  distance  from  the  centre. 

A  straight  line  drawn  from  the  centre  to  the  circumference 
is  called  a  radius  (plural,  radii). 

A  straight  line  drawn  through  the  centre  and  extending  to 
the  circumference  in  both  directions  is  called  a  diameter. 

A  straight  line  joining  any  two  points  of  a  circumference 
is  called  a  chord. 

A  part  of  a  circumference  is  called  an  arc. 

Every  diameter  divides  the  circle  into  two  equal  parts, 
called  semicircles.  Every  diameter  divides  the  circumference 
into  two  equal  parts,  called  semicircumferences. 

EXERCISES 

1.  Name  in  Fig.  45  a  radius,  a  chord,  an  arc,  a  diameter. 

2.  What  is  true  of  all  radii  of  the  same  circle  ? 

3.  What  is  true  of  all  diameters  of  the  same  circle  ? 

4.  Compare  a  radius  with  a  diameter  as  regards  lengih. 

5.  Compare  a  diameter  and  a  chord  as  regards  length. 

6.  In  Fig.  45,  what  is  described  by  any  point  in  the  line  OA  ? 

7.  What  is  represented  by  the  tire  of  a  carriage  wheel?  by  the 
spokes?    by  the  part  of  the  tire  between  two  spokes? 

8.  Can  you  name  any  objects  that  are  circular  in  shape  ? 


32 


FIRST   STEPS   IN   GEOMETRY 


30.    Compasses.     Circles  are  described  on  paper  by  means 
of  compasses  (Fig.  46).     The  dividers  (Fig.  10,  p.  6)  can  be 

changed  to  compasses  by  substitut- 
ing for  one  of  the  metal  points  a 
pen  or  a  pencil. 

The  compasses  should  be  held 
between  the  thumb  and  the  fore- 
finger, as  shown  in  Fig.  46,  and  the 
metal  point  should  be  pressed  against 
the  paper  just  hard  enough  to  keep 
it  in  place. 

On  the  blackboard,  circles  may  be 
described  with  the  aid  of  a  string. 
Make  a  loop  at  one  end  of  the  string  and  slip  it  over  the 
end  of  a  crayon.  Press  the  string  against  the  blackboard  at 
the  point  chosen  for  the  centre.  Keep  the  string  stretched 
while  you  move  the  crayon  round  the  centre. 


Fig.  46. 


Fig.  47.  — How  Circles  are  made  on  the  Ground. 


In  gardens,  parks,  etc.,  it  is  sometimes  desirable  to  mark 
out  large  circles.  The  method  by  which  circles  are  com- 
monly drawn  in  gardens  is  illustrated  in  Fig.  47. 


GEOMETRIC    MAGNITUDES    AND   MOTION 


33 


31.    The  Cylinder,  the  Cone,  and  the  Sphere 


A  straight 
axis  about  which  it  revolves 
In  any  other  position  the  line 


line  perpendicular  to  the 
generates  a  plane  surface, 
generates  a  curved  surface. 

When  a  rectangle  ABCD  (Fig.  48)  revolves  about  one  side 
AD  as  an  axis,  the  opposite  side  BC  generates  the  curved 
surface  of  a  solid  called  a  cylinder. 

When  a  right  triangle  ABC  (Fig.  49)  revolves  about  one  of 
the  sides  that  form  the  right  angle,  the  side  opposite  the  light 
angle  generates  the  curved  surface  of  a  solid  called  a  cone. 


Fig.  48. 


Fig.  49 


When  a  semicircumference  ABC  (Fig.  50)  revolves  about 
the  diameter  AC  as  an  axis,  it  generates  the  curved  surface 
of  a  solid  called  a  sphere. 

EXERCISES 

1.  What  is  generated  by  the  sides  AB  and  DC  oi  the  rectangle  in 
Fig.  48?     What  is  generated  by  the  points  B  and  C? 

2.  What  is  generated  by  the  side  BC  of  the  triangle  in  Fig.  49? 
What  is  generated  by  a  point  of  ^J5  between  A  and  J5? 

3.  What  is  generated  by  the  point  B  on  the  curve  ABC  in  Fig.  50, 
as  the  curve  revolves  about  the  axis  ^IC? 

4.  What  solid  is  generated  when  a  rectangular  piece  of  cardboard  is 
moved  in  a  straight  line  perpendicular  to  its  own  plane  ? 

5.  How  can  a  plane  surface  be  moved  so  as  not  to  generate  a  solid  ? 


34  FIRST   STEPS    IN   GEOMETRY 

32.  Drawing  Exercises.  These  exercises  require  the  use  of 
the  instruments  hitherto  mentioned,  and  also  a  rubber  eraser  to 
remove  auxiliary  lines,  or  lines  which  aid  in  constructing  the 
figure  but  do  not  belong  to  the  figure  when  completed. 

If  for  any  reason  you  wish  to  leave  auxiliary  lines  in  the 
figure,  they  should  be  drawn  with  short  dashes. 

EXERCISES 

1.  Describe  a  circle  with  a  radius  of  1  in. 

2.  Describe  a  circle  with  a  radius  of  3  cm. 

3.  Describe  a  circle  and  then  draw  a  chord  equal  to  the  radius.  Also 
draw  a  chord  equal  to  twice  the  radius. 

4.  Describe  three  concentric  circles,  that  is,  circles  having  the  same 
point  as  centre. 

5.  Describe  two  circles  such  that  the  circumference  of  each  shall  pass 
through  the  centre  of  the  other.     How  do  they  compare  in  size  ? 

6.  Describe  three  unequal  circles  so  that  the  centre  of  each  shall  lie 
on  the  circumference  of  one  of  the  other  two. 

7.  Describe  four  equal  circles  such  that  the  circumference  of  each 
circle  shaU  touch  but  not  cut  the  circumferences  of  two  of  the  other 
circles. 


Fig.  51.  Fig.  52. 

8.    Construct  Fig.  51.     The  circle  and  the  arcs  have  the  same  radius. 
Then  make  a  six-rayed  leaf  (Fig.  52)  by  erasing  the  circumference. 


GEOMETRIC   MAGNITUDES   AND   MOTION 


35 


9.  Construct  an  equilateral  triangle,  or  triangle  having  three  equal 
sides  (Fig.  53). 

Draw  a  straight  line,  AB.  With  A  and  B  as  centres,  and  the  length 
AB  Z.S  radius,  describe  arcs  intersecting  at  C.  Join  by  straight  lines 
A  to  C,  and  B  to  C.     ABC  \^  an  equilateral  triangle. 

Erase  the  arcs  when  C  has  been  found. 


Fig.  53. 


Fig.  54. 


10.    Construct  a  trefoil  (Fig.  54). 

Make  an  equilateral  triangle.  Find  by  trial  the  middle  point  of 
one  side.  Describe  three  arcs  of  circles  with  the  corners  of  the  triangle 
as  centres,  and  a  radius  equal  to  half  the  side  of  the  triangle. 


Fig.  55. 


Fig.  56. 


11.    Construct  the  ornamental  design  shown  in  Fig.  56. 

First  make  a  square  as  in  Fig.  55,  and  join  the  opposite  corners. 

From  the  four  corners  as  centres  draw  arcs  with  the  length  of  one 
side  of  the  square  as  a  radius  ;  and  also  from  the  four  corners  as  centres 
draw  the  four  other  arcs  with  a  smaller  radius. 

Then  erase  lines  so  as  to  show  Fig.  56. 


36  FIRST   STEPS   IN   GEOMETRY 

Construct  the  ornamental  figures  given  below : 


Fig.  57. 


12.  Construct  equal  adjacent  squares  (Fig.  57),  and  draw  semicircles 
on  the  sides  of  the  squares  as  diameters.  Then  erase  lines  so  as  to 
show  the  figure  which  appears  on  the  right. 


Fig.  58.  Fig.  59. 

13.  Construct  a  square,  draw  two  concentric  circles  with  the  radius 
of  the  smaller  circle  two  thirds  that  of  the  larger,  and  draw  the  semi- 
circles, as  shown  in  Fig.  58.     Then  erase  lines  so  as  to  show  Fig.  69. 


Fig.  60. 


14.  Draw  the  indicated  circles  (Fig.  60),  with  the  radius  of  the  larger 
ones  twice  that  of  the  smaller.  Then  erase  lines  so  as  to  show  the 
figure  as  it  appears  on  the  right. 


E 

D, 

%. 

A 

C 

B 

GEOMETRIC    MAGNITUDES   AND   MOTION  37 

33.  Loci.  When  a  point  describes  the  circumference  of  a 
circle  it  has  to  obey,  so  to  speak,  the  law  that  it  must  always 
he  at  the  same  distance  from  the  centre  of  the  circle. 

In  general,  the  path  of  a  point  mov- 
ing in  obedience  to  a  prescribed  law  is 
called  the  locus  of  the  point.  The 
plural  of  locus  is  loci. 

Let  us  consider  some  more  examples 
of  loci. 

Suppose  that  a  point  starts  at  C 
(Fig.  61),  halfway  between  two  fixed 
points  A  and  B^  and  describes  a  straight  ^^^''  ^^' 

line  CD  perpendicular  to  the  line  AB.  The  moving  point  will 
always  he  equally  distant  from  A  and  B.  For,  let  us  take  any 
point  D  in  the  perpendicular,  and  let  us  imagine  that  the 
figure  ACD  is  folded  about  CD  through  half  a  revolution. 
The  line  CA  will  fall  upon  the  line  CB^  because  the  right 
angles  at  C  are  equal ;  the  point  A  will  fall  upon  the  point 
i?,  because  CA  =  CB ;  and,  therefore,  the  line  DA  will  coin- 
cide with  the  line  DB,  for  only  one  straight  line  can  be  drawn 
between  the  two  points  D  and  B.     Therefore,  DA  =  DB. 

On  the  other  hand,  any  point  ^  not  in  the  perpendicular 
CD  is  unequally  distant  from  A  and  B ;  for  JEA  =  UD  -f  DA 
=  £JD  +  DB  (since  DA  and  DB  are  equal),  and  is,  there- 
fore, greater  than  JSB ;  for  a  straight  line  is  the  shortest  line 
between  two  points.     Hence, 

The  locus  of  a  point  so  moving  in  a  plane  that  it  is  always 
equidistant  from  two  fixed  points  in  that  plane  is  the  perpen- 
dicular erected  at  the  middle  of  the  straight  line  which  joins 
the  two  fixed  points. 

The  perpendicular  DC  divides  the  straight  line  AB  into  two 
equal  parts,  and  is  called  the  perpendicular  bisector  of  AB. 


38 


FIRST    STEPS    IN    GEOMETRY 


Fig.  62. 


34.  The  Ellipse.  Fasten  a  sheet  of  paper  to  a  smooth  board 
by  two  pins  at  F  and  F^  (Fig.  62).     Put  over  the  pins  a  string 

in  the  form  of  a  loop,  taking  care 
that  the  loop  is  somewhat  longer 
than  the  distance  FF'.  Then  move 
the  point  P  of  a  pencil,  keeping  it 
pressed  against  the  string,  as  shown 
in  Fig.  62. 

The  point  P  will  trace  a  smooth 

closed  curve  called  an  ellipse.     Tlie 

points  F  and  F^  are  called  the  foci  of 

the  ellipse. 

The  distance  PF  +  PF'  remains  the  same  whether  P  is 

in  motion  or  at  rest. 

An  ellipse  is  the  locus  of  a  point  so  moving  that  the  sum  of 
its  distances  from  two  fixed  points  (the  foci)  is  constant. 

The  length  ^^'  is  equal  to  PF  +  PF^  and  is  called  the 
major  axis  of  the  ellipse. 

The  major  axis  bisects  the  curve  and  the  area  enclosed 
by  the  curve. 

The  less  the  distance  between  the  foci  the  more  nearly  the 
form  of  an  ellipse  approaches  that  of  a  circumference  of  a 
circle.  If  the  foci  coincide,  in  other  words,  if  only  one  pin 
is  used,  the  curve  will  be  a  circumference. 

If  you  stand  in  front  of  a  luminous  point,  revolving  about  an 
axis,  it  will  appear  to  describe  a  circumference.  As  you  move 
sideways  the  curve  will  appear  to  change  to  an  ellipse  more 
and  more  oval  in  shape,  and  finally  appear  as  a  straight  line. 
The  ellipse  is  a  curve  often  used  in  cabinet  work  (mirrors, 
tables,  etc.),  in  the  arches  of  bridges,  and  in  architecture. 
The  earth  and  all  the  other  planets  move  round  the  sun  in 
elliptical  curves,  the  sun  being  at  one  of  the  foci. 


GEOMETRIC   MAGNITUDES   AND   MOTION  39 

35.    The  Parabola.     Let  a  point  F  (Fig.  63)  and  a  straight 
line  CE  be  fixed  in  position.      Draw  FD  perpendicular  to 
CE,     Then  FD  is  the  shortest  line  that  can 
be  drawn  from  F  to  CE^  and  is  called  the 
distance  from  F  to  CE. 

Now,  suppose  a  point  starts  at  A^  half- 
way between  F  and  Z>,  and  so  moves  that  ^ 
it  is  always  equidistant  from  F  and  CE. 

The  point  cannot  move  from  ^  in  a 
straight  line ;  for  there  can  be  no  straight 
line  every  point  of  which  is  equidistant 
from  the  fixed  point  F  and  the  fixed  straight  ^^^-  ^• 

line  CE. 

The  point  cannot  describe  a  circumference  about  F  as 
centre ;  for  in  that  case  its  distance  from  CE  would  vary, 
while  its  distance  from  F  would  not  change. 

The  actual  path  of  the  point  is  a  curve  PAQ^  which  is 
called  a  parabola. 

A  parabola  is  the  locus  of  a  point  so  moving  that  it  is  always 
equidistant  from  a  given  point  and  a  given  straight  line. 

The  line  AF  extended  is  called  the  axis  of  the  parabola,  and 
the  point  F  is  called  the  focus  of  the  parabola.  The  parabola 
evidently  extends  towards  the  right,  both  above  and  below 
the  axis,  as  far  as  the  point  is  supposed  to  move. 

The  portion  of  the  parabola  above  the  axis  is  exactly 
similar  to  the  portion  below  the  axis. 

The  parabola  is  a  very  graceful  curve.  The  path  of  a  ball, 
when  the  ball  is  thrown  in  any  other  direction  than  a  vertical 
one,  is  very  nearly  a  parabola.  The  same  is  true  of  a  jet  of 
water  issuing  from  a  pipe  in  a  horizontal  or  inclined  direction. 

If  a  parabola  is  revolved  about  its  axis,  it  generates  a  sur- 
face called  a  parabolic  surface. 


40 


FIRST   STEPS   IN    GEOMETRY 


A  parabolic  surface  made  of  metal  polished  on  the  concave 
side  is  called  a  parabolic  reflector.  When  a  light  is  placed 
at  the  focus,  all  the  rays  of  light  that  fall  on  the  reflector 
are  reflected  in  straight  lines  parallel  to  the  axis  (Fig.  64). 


Fig.  64. 


Advantage  is  taken  of  this  property  in  the  construction  of  the 
headlights  of  locomotives  for  the  purpose  of  concentrating  the 
light  and  throwing  it  as  far  as  possible  along  the  track. 


Fig.  65. 


Brooklyn  Suspension  Bridge. 


The   steel   cables   that  support   a  suspension    bridge    are 
parabolic  curves  (Fig.  65). 


GEOMETRIC   MAGNITUDES   AND   MOTION 


41 


36.  The  Cycloid.  Consider  the  motion  of  a  point  P 
(Fig.  %Q)  on  tlie  rim  of  a  wheel,  as  the  wheel  rolls  along  a 
straight  track. . 


K 

Fig.  66.  —The  Cycloid. 


If  the  wheel  simply  revolved  about  its  centre  C,  the  point 
P  would  describe  a  circumference ;  if  the  wheel  did  not 
revolve,  but  was  made  to  slide  along  the  track,  the  point  P 
would  describe  a  straight  line  parallel  to  OX.  But  the  wheel 
has  both  motions ;  it  revolves  about  C  and  also  moves  along 
the  track.  The  consequence  is  that  the  point  P  describes  a 
curve  OPHX^  which  is  called  a  cycloid. 

A  cycloid  is  the  locus  of  a  point  moving  on  the  circumference 
of  a  circle  as  the  circle  rolls  along  a  straight  line. 

The  length  of  the  cycloid  from  0  to  X  is  exactly  four 
times  the  diameter  of  the  generating  circle. 

The  area  of  the  surface  between  the  curve  OHX  and  the 
straight  line  OX  is  exactly  three  times  the  area  of  the 
generating  circle. 

37.  The  number  of  differeiit  curved  lines  that  may  be 
generated  by  points  moving  according  to  different  laws  is 
unlimited.  But  in  elementary  geometry  the  straight  line  and 
the  circumference  of  the  circle  are  the  only  lines  selected 
for  study. 


42 


FIRST    STEPS   m    OPTOMETRY 


38.  Generation  of  Angles.  An  angle  has  already  been  defined 
(p.  9).  Let  us  now  regard  an  angle  as  a  magnitude  generated 
by  motion.  When  a  straight  line  OA  (Fig.  67)  describes  a 
circle  by  revolving  about  the  point 
0,  it  also  generates  an  angle  of  con- 
stantly increasing  magnitude.  One 
side  of  the  angle  is  OA  in  its  first 
position;  the  other  side  is  OB,  OC, 
etc.  If  OD  is  perpendicular  to  OA, 
the  angle  generated  AOD  is  a  right 
angle.  When  the  revolving  line 
reaches  the  position  OF,  drawn  oppo- 
site to  OA,  the  angle  generated  is 
equal  to  two  right  angles,  and  is  called  a  straight  angle. 
When  the  revolving  line  has  arrived  at  the  position  OG,  the 
angle  generated  AOG  m  equal  to  three  right  angles.  When 
the  revolving  line  arrives  at  its  original  position  OA,  it  has 
made  one  revolution  and  generated  an  angle  equal  to  four 
right  angles,  or  two  straight  angles. 

An  angle  less  than  a  right  angle  is  called  an  acute  angle. 
An  angle  greater  than  a  right  angle  and  less  than  a  straight 
angle  is  called  an  obtuse  angle. 

The  angle  formed  by  any  two  straight  lines  may  be  regarded 
as  generated  by  one  of  the  lines  turning  about  the  vertex  till 
it  coincides  with  the  other  line.  Hence,  the  magnitude  of 
an  angle  depends  only  on  the  amou?it  of  rotation  required  to 
generate  the  angle. 

The  common  units  for  measuring  angles  are  : 

the  degree  (°),  the  minute  ('),  the  second  (''). 

A  right  angle  is  equal  to  90°,  a  degree  is  equal  to  60',  and 
a  minute  is  equal  to  60''. 


GEOMETRIC    MAGNITUDES   AND   MOTION 


43 


Two  angles  are  called  supplementary  angles,  if  their  sum  is 
180° ;   and  each  is  called  the  supplement  of  the  other. 

Two  angles  are  called  complementary  angles,  if  their  sum  is 
90°  ;   and  each  is  called  the  complement  of  the  other. 


Fig.  G8. 


EXERCISES 

1.  How  many  angles  are  formed  by  the  two  intersecting  straight 
lines  in  Fig.  68?     How  many  acute  angles? 
How  many  obtuse  angles?      Are  the  acute 
angles  equal?     Are  the  obtuse  angles  equal? 

2.  What  ki-nd  of  an  angle  do  the  hands 
of  a  clock  make  at  2  o'clock?  3  o'clock? 
4  o'clock?    6  o'clock? 

3.  Mention  the  hours  when  the  hands  of  a  clock  form  a  right  angle, 
an  acute  angle,  an  obtuse  angle. 

4.  In  what  period  of  time  is  a  right  angle  described  by  the  minute 
hand  ?  by  the  hour  hand  ? 

5.  How  many  degi-ees  does  the 
hour  hand  describe  in  1  hour?  in  3 
hours  ?  in  5  hours  ?  in  24  hours  ? 

6.  By  what  angle  must  a  man 
walking  north  change  the  direction  of 
his  motion  in  order  to  walk  towards  the 
east  (Fig.  69)  ? 

In  order  to  walk  towards  the  south? 
In  order  to  walk  N.E.  ? 
In  order  to  walk  S.E.  ? 
In  order  to  walk  S.W.  ? 

7.  How  many  degrees  are  there  in  four  right  angles? 

8.  How  many  minutes  are  there  in  a  right  angle  ? 

9.  Reduce  10°  6'  20''  to  seconds,  and  18,000''  to  degrees. 

10.  Find  the  supplements  of  30°,  45°,  60°,  80°,  120°,  150°. 

11.  Find  the  complements  of  10°,  20°,  30°,  40°,  60°,  75°. 

12.  What  angle  has  a  supplement  equal  to  itself? 

13.  What  angle  has  a  complement  equal  to  itself  ? 

14.  What  angle  has  a  complement  equal  to  half  of  itseK? 


44 


FIRST   STEPS   IN    GEOMETRY 


39.  Measurement  of  Angles.  A  practical  method  for  the 
measurement  of  angles  is  obtained  by  taking  advantage  of 
a  simple  relation  between  an  angle  and  the  arc  between  its 
sides,  described  by  the  revolution  of  one  side  of  the  angle 
about  the  vertex  till  it  coincides  with  the  other  side.  The 
angle  and  the  arc  increase  at  the  same  rate.  If  the  angle  is 
doubled,  the  arc  is  doubled ;  if  the  angle  is  trebled,  the  arc 
is  trebled,  and  so  on. 

Suppose  (Fig.  70)  that  the  angles  AOB^  BOC  are  equal. 
Then  the  arcs  AB  and  BC  must  also 
be  equal.  For,  if  we  fold  A  OB  about 
OB  as  an  axis  through  half  a  revolu- 
tion, OA  will  fall  upon  0(7,  because 
the  angles  BOA  and  BOC  are  equal ; 
and  A  will  fall  on  C,  because  OA  and 
OC  are  equal  radii ;  therefore,  the  arc 
BA  will  coincide  with  the  arc  BC,  and 
be  equal  to  it  in  length.  Hence, 
Equal  angles  at  the  centre  of  a  circle 
intercept  equal  arcs  on  the  circumference. 

This  truth  is  applied  to  the  measurement  of  angles  by 
dividing  arcs  into  degrees,  minutes,  and  seconds,  precisely  as 
angles  are  divided.  An  entire  circumference,  therefore,  is 
divided  into  360  equal  parts  called  degrees.  Then,  to  find 
the  number  of  degrees,  minutes,  and  seconds  in  an  angle,  we 
have  only  to  find  the  number  of  degrees,  minutes,  and  seconds 
in  an  arc  described  from  the  vertex  of  the  angle  as  a  centre, 
and  contained  between  its  sides.  Hence,  we  say  that  an 
angle  at  the  centre  of  a  circle  is  measured  hy  its  intercepted 
arc,  meaning  that  the  angle  contains  as  many  angle  degrees, 
minutes,  and  seconds  as  its  intercepted  arc  contains  arc  degrees, 
minutes,  and  seconds. 


GEOMETRIC   MAGNITUDES   AND  MOTION 


45 


40.    The  Protractor.     On  paper,  angles  are  measured  and 
constructed  with  the  aid  of  an  instrument  called  a  protractor. 


Fig.  71.  —  A  Protractor, 


A  circular  protractor  (Fig.  71)  consists  of  a  semicircular 
piece  of  cardboard,  horn,  or  metal.  The  arc  is  divided  into 
degrees.  In  protractors  made  of  horn  or  metal  an  open  space 
is  left  round  the  centre  to  make  easier  the  construction  or 
measurement  of  angles  with  short  sides. 

To  measure  an  angle,  place  the  centre  of  the  protractor  on 
the  vertex  of  the  angle,  and  make  the  diameter  of  the  pro- 
tractor coincide  with  one  side  of  the  angle ;  then  read  on  the 
divided  edge  of  the  protractor  the  division  through  which  the 
other  side  of  the  angle  passes. 

To  construct  an  angle  of  given  value,  draw  a  straight  line, 
place  the  protractor  so  that  its  diameter  coincides  with  the 
line,  mark  the  place  of  the  centre  of  the  protractor,  and  also 
the  point  where  an  arc  containing  the  given  number  of 
degrees  ends.  Then  draw  a  straight  line  through  the  points 
marked. 

Protractors  are  sometimes  made  in  the  shape  of  a  rectangle. 


46 


FIRST   STEPS   m   GEOMETRY 


Fig.  72. 


EXERCISES 


1.  Make  with  a  protractor  an  angle  of  30°. 

2.  Make  with  a  protractor  an  angle  of  45°. 

3.  Make  with  a  protractor  an  angle  of  100°. 

4.  Make  with  a  protractor  an  angle  of  150°. 

5.  Draw  an  angle  and  a  straight  line.  Then  constmct  upon  the 
straight  line  as  one  side,  with  the  aid  of  a  protractor,  an  angle  equal  to 
the  angle  you  have  drawn. 

6.  Make  the  angles  30°,  45°,  and  60°  as  well  as  you  can  by  ruling 
the  lines,  but  estimating  the  angle  by  the  eye.  Then  measure  the 
angles  and  report  the  error  for  each  case. 

7.  Construct  an  equilateral  triangle  (see  p.  35)  and  measure  its 
angles.     What  is  the  sum  of  the  three  angles  ? 

8.  Draw  a  right  triangle,  measure  the  two  smaller  angles,  and  find 
their  sum. 

9.  Draw  a  three-sided  figure  with  unequal  sides,  measure  its  angles, 
and  find  their  sum. 

10.  Draw  a  four-sided  figure  with  unequal  sides,  measure  its  angles, 
and  find  their  sum. 

11.  Draw  a  four-sided  figure  with  equal  sides,  measure  its  angles, 
and  find  their  sum. 


GEOMETRIC    MAGNITUDES    AND   MOTION  47 

41.  Axial  Symmetry.  If  we  draw  a  perpendicular  DC 
from  a  point  D  (Fig.  73)  to  a  straight  line  AB  and  extend  it 
to  D'  making  CD'  equal  to  (7Z>,  the  points  D  and  I>'  are  said  to 
be  symmetric  points  with  respect  to  the 
line  AB,      Suppose  that  CD  is  made 


to  revolve  about  AB  through  half  a     V  ^ 

revolution.      During  this  motion   CD       \ 
will  not  change   in  length,  and  will        ^ 
remain  perpendicular  to  AB.     There- 
fore, after  half  a  revolution  CD  will 
fall  upon   CD\   and  D    will  coincide 
with  D'.     Accordingly,  D  and  D'  are  B 

said  to  have  axial  symmetry  with  respect  fi<*-  '^^■ 

to  the  line  vJ/>,  and  AB  is  called  the  axis  of  symmetry. 

The  points  E  and  E'  are  also  symmetric  points  with  respect 
to  the  line  AB. 

Any  two  points  so  situated  that  the  straight  line  joining  them 
is  perpendicular  to  AB  and  bisected  by  AB  are  symmetric  points 
ivith  respect  to  AB. 

The  straight  lines  DE  and  D' E'  are  symmetric  lines  with 
respect  to  the  axis  AB.  By  folding  DE  about  AB  we  can 
make  it  coincide  with  D'E\  and  thus  see  that  for  every  point 
in  DE  there  is  a  corresponding  symmetric  point  in  D'E'. 

The  plane  figures  CDEF  and  CD'E'F  are  symmetric  figures 
with  respect  to  the  axis  AB.     In  general. 

Two  plane  figures  are  said  to  be  symmetric  with  respect  to  an 
axis  if  every  point  in  the  boundary  of  one  has  a  corresponding 
symmetric  point  in  the  boundary  of  the  other. 

Two  plane  figures  symmetric  with  respect  to  an  axis,  as 
CDEF  and  CD'E'F,  are  equal  figures  ;  for  either  can  be  made 
to  coincide  with  the  other  by  revolving  it  about  AB  through 
half  a  revolution. 


48 


FIRST   STEPS   IN    GEOMETRY 


The  single  figure  DEE'D'  (Fig.  73,  p.  47),  formed  by  putting 
together  the  equal  figures  CDEF  and  CD'E'F^  is  sometimes 
called  a  symmetric  figure  with  respect  to  AB^  but  a  better 
name  for  it  is  a  bi-symmetric  figure,  having  AB  for  the  axis 
of  symmetry.  Therefore,  a  plane  figure  is  bi-symmetric  if 
it  can  be  divided  by  a  straight  line  into  two  equal  parts, 
symmetric  with  respect  to  the  line  of  division. 

The  bi-symmetric  form  is  not  confined  to  geometry.  It  is 
seen  in  the  shapes  of  various  common  objects,  such  as  a  bottle 
or  a  lamp  shade.  It  is  employed  with  numerous  variations 
in  the  decorative  arts  and  in  architecture.  It  is  the  form 
chosen  by  nature  for  the  shape  of  a  leaf. 


EXERCISES 

1.  Name  a  bi-symmetric  four-sided  figure 
in  Fig.  73,  p.  47. 

2 .  Name  two  symmetric  triangles  in  Fig.  7  3 . 

3.  Compare  the  angles  which  two  symmet- 
rically placed  straight  lines  form  with  the  axis 
of  symmetry.  Illustrate  by  referring  to 
Fig.  73. 

4.  In  Fig.  71  name  the  axis  of  symmetry, 
pairs  of  corresponding  points,  pairs  of  sym- 
metric lines,  two  symmetric  figures,  and  a 
bi-symmetric  figure. 

5.  Make  a  bi-symmetric  figure  with  paper  and  scissors. 
Draw  on  the  paper  a  straight  line  AB  (Fig.  75). 

Fold  the  paper  on  AB  and  sketch  lightly  the  lines  AC 
and  CB  on  the  upper  fold.  Then  cut  out  the  figure  along 
the  lines  AC  and  CB.  Unfold,  and  the  bi-symmetric 
figure  is  formed,  as  seen  on  the  right  (Fig.  75). 

6.  Make  with  paper  and  scissors  the  bi-symmetric 
figure  shown  on  the  right  in  Fig.  76. 

7.  Make    in    a   similar  way   a  bi-symmetric  figure 
according  to  your  own  design. 


Fig.  76. 


GEOMETRIC    MAGNITUDES   AND   MOTION 


49 


h^ 


"-"c 


Fig.  77. 


8.  Must  a  figure  made  in  this  way  always  be  bi-symmetric  ? 

9.  Can  symmetric  figures  be  bounded  by  curved  lines  ? 

10.  Make  a  cross  like  that  shown  in  Fig.  77. 
Begin  by  drawing  carefully  one  half  the  cross  as 

represented  in  the  figure.  Take  the  lengths  as  fol- 
lows :  AB,  4  in. ;  AF,  ^  in.;  FG,  GE,  and  ED,  1  in. 
Fold  on  AB  as  an  axis.  Prick  pin  holes  through 
the  points  F,  G,  etc.  Unfold,  and  connect  the  pin 
holes  on  the  left  of  AB  by  straight  lines. 

11.  Make  a  spool  like  that  shown  in  Fig.  78  by  the  pin-hole  method. 

12.  Make  the  spool  in  Fig.  78  by  the  printing  method. 
First  construct  the  right  half  of  the  spool,  using  a 

soft  lead  pencil.  Moisten  the  paper  with  a  sponge. 
Then  fold  on  the  axis,  lay  a  book  over  the  paper,  and 
strike  it  a  blow  with  the  fist.     Then  unfold. 

"ISv   Devise  a  way  of  constructing  the  bi-symmetric 
figure  in  Fig.  77  or  in  Fig.  78  without  folding  the  paper, 
struct  the  figure. 

14.  Construct  the  bi-symmetric  figure  suggested 
in  Fig.  79,  A. 

15.  Construct  the  bi-symmetric  figure  suggested 
in  Fig.  79,  B. 

16.  Construct  the  bi-symmetric  figure  suggested 
in  Fig.  79,  C. 

17.  Is  a  sorrel  leaf  (Fig.  ^0,  A)  a  bi- 
symmetric  figure?  Give  a  reason  for  your 
answer.  Is  the  lorm  in  Fig.  80,  B,  bi- 
symmetric  ?     Construct  it. 

18.  Draw  a  square,  and  then  draw  as 
many  axes  of  symmetry  as  possible. 

19.  Draw  a  rectangle,  and  then  draw  as 
many  axes  of  symmetry  as  possible. 

20.  Draw  an  equilateral  triangle,  and 
then  draw  as  many  axes  of  symmetry  as 
possible. 

21.  Point  out  the  bi-symmetric  objects 
in  the  room. 


Fig.  80. 


50 


FIRST   STEPS   IN   GEOMETRY 


42.  Central  Symmetry.  If  a  plane  figure  after  being 
turned  in  its  own  plane  about  a  point  through  half  a  revolu- 
tion coincides  with  its  original  position,  tlie  figure  is  said  to 
have  twofold  symmetry  with  respect  to  the  point ;  and  the  point 
is  called  the  centre  of  symmetry. 

An  example  of  twofold  symmetry  is  seen  in  Fig.  81. 

In  the  figure  ABCD  the  lines  AC  and  BD^  intersecting  at 
0,  are  perpendicular  to  each  other.  Also  OA  =  0C\  and 
OB  =  OD.     Pin  tracing  paper  upon  the  figure  at  0,  and  trace 


Fig.  82. 


its  outline.  Then  turn  the  tracing  paper  halfway  round 
the  pin  as  an  axis.  A  and  C  exchange  places;  likewise  B 
and  D  ;  and  the  two  figures  again  coincide. 

If  we  regard  Fig.  81  as  an  example  of  axial  symmetry,  how 
many  axes  of  symmetry  has  it  ?     Name  them. 

If  a  figure  coincides  with  its  original  position  after  being 
turned  one  third  of  a  revolution  in  its  own  plane  about  a 
point,  the  figure  is  said  to  have  threefold  symmetry,  and  so  on. 

An  equilateral  triangle  (Fig.  82)  has  threefold  symmetry 
with  respect  to  the  intersection  0  of  the  perpendiculars  drawn 
from  the  vertices  to  the  opposite  sides,  as  may  be  shown  by 
the  aid  of  tracing  paper. 

If  we  regard  Fig.  82  as  an  example  of  axial  symmetry,  how 
many  axes  of  symmetry  has  it  ?     Describe  their  situation. 


GEOMETRIC   MAGNITUDES   AND   MOTION 


51 


Nature  shows  a  marked  preference  for  forms  having  central 
symmetry.     This  is  seen  by  examining  ^ 

the  petals  of  flowers  (Fig.  83)  and  the  ^^^ 

forms  of  snow  crystals  (Fig.  85,  i\  G,  H). 

EXERCISES 

1.  Make  petals  similar  to  those  of  which 
the  'flowers  in  Fig.  83  are  composed,  but  having 
perfect  central  symmetry. 

Take  thick  paper,  and  describe  upon  it  a  cir- 
cle. Divide  the  circumference  into  three  equal 
parts.  Draw  radii  to  two  of  the  points  of  divi- 
sion.    Between  the  radii  sketch  free-hand  one 

of  the  petals.     Cut  out 

the  figure.     This  figure 

is  a  pattern  of  one  of  the 

petals.     Place  it  on  the  paper. 

Pass  a  pin  through  the  point  of  the  petal  and  turn 

the  pattern  about  the  pin  as  a  centre  until  you 
can  again  draw  its  outline  beside  the  first  sketch.  Repeat,  and  Fig.  84 
will  be  the  result. 


/ 


Fig.  83. 


Draw  its  outline. 


Fig.  84. 


Fig.  85. 


2.    How  many-fold  symmetry  has  each  of  the  figures  in  Fig.  85  ? 


52  FIRST    STEPS   IN    GEOMETRY 

43.  Distance  from  a  Point  to  a  Straight  Line.  The  shortest 
line  from  a  point  to  a  straight  line  is  the  perpendicular  from 
the  point  to  the  line. 

This  truth  may  be  established  by  the  aid  of  axial  symmetry 
as  follows  : 

Let  FC  (Fig.  86)  be  a  perpendicular  drawn  from  a  point  P 
to  a  straight  line  AB^  and  let  FD  be  any  other  straight  line 
p  drawn  from  F  to  AB.     Extend  FC  to  F'  mak- 

ing CF'  equal  to  FC,  and  draw  F^F.     The 
triangles  FFC  and  P'Z) (7  are  symmetric  with 

respect  to  AB.     Therefore,  they  are  equal, 

^\      i^'  and  FD  =  F'D.     Now  FF'  is  less  than  FD 

\  \  H-  P'P,  because  a  straight  line  is  the  shortest 

\  ^  line  from  one  point  to  another.     But  FF'  = 

\^   T.  2 PC,    and   FD  +  F'D  =2FD.      Therefore, 

2  PC  is  less  than  2  PP.     Hence,  it  follows, 

by  dividing  by  2,  that  PC  is  less  than  PP. 

This  is  true  however  FD  is  drawn,  provided  it  is  not  per- 
pendicular to  AB.  Hence,  we  conclude  that  the  shortest  line 
from  P  to  AB  is  the  perpendicular  FC. 

The  length  of  the  perpendicular  from  a  point  to  a  straight 
line  is  called  the  distance  from  the  point  to  the  line. 

44.  Parallel  Lines  further  examined.  Parallel  lines  have 
been  defined  as  straight  lines  that  lie  in  the  feame  plane  and 
cannot  meet  however  far  extended  (p.  11). 

Now  it  is  easy  to  imagine  two  straight  lines  as  extending 
indefinitely,  but  our  power  to  observe  and  represent  them  is 
very  limited,  and  practically  confined  to  comparatively  short 
distances.  How  then  can  we  be  sure  tiiat  parallel  lines  really 
exist  ?  This  question  may  be  answered  by  again  calling  axial 
symmetry  to  our  aid. 


GEOMETRIC    MAGNITUDES   AND   MOTION  53 

Let  AB  and  CD  (Fig.  87)  be  two  straight  lines  perpendicu- 
lar to  the  straight  line  JfiV,  and  intersecting  it  at  the  points 
E  and  F^  respectively.    The  figures  3^ 

AEFC  and  BEFD  are  symmetric 


-D 


with  respect  to  MN  as   an  axis.  ^ 

Hence,  if  AEFC  is  revolved  about 

MN^  it  can  be  made  to  coincide 

with  BEFD.      Therefore,    if   the  ^ 

lines  AB  and  CD  intersect  on  the  ^^'  *^* 

left  of  JfJV,  they  must  also  intersect  on  the  right  of  MN\ 

that  is,  they  must  have  two  points  of  intersection.     Now  two 

straight  lines  can  have  only  one  point  of  intersection.     We 

conclude,  therefore,  that  the  two  lines  cannot  meet  at  all. 

Two  straight  lines  in  a  plane  perpendicular  to  the  same 
straight  line  are  parallel  lines. 

The  line  MN  is  perpendicular  to  both  the  parallels  AB  and 
CD.  It  can  be  proved  that  through  any  point  on  one  of  two 
parallels  a  straight  line  perpendicular  to  both  can  be  drawn. 
The  length  of  this  common  perpendicular  comprised  between 
the  parallels  is  taken  as  the  distance  between  the  parallels  at 
that  point. 

From  axial  symmetry  it  follows  that 

Two  parallel  straight  lines  are  everywhere  equally  distant. 

Let  AB  and  CD  (Fig.  88)  be  perpendicular  to  MN^  and 

therefore  parallel  lines.      Let 

M 

^  Q         EF  be  a  common  perpendicular 


'^  to  AB  and  CD,  and  GIf  be  the 

j^ — D  position  taken  by  EF  after  it 

^  is  revolved  about  MN  through 


Fig.  88. 


half  a  revolution.  From  sym- 
metry it  follows  that  EF  =  GH,  and  this  equality  holds  good 
wherever  on  the  line  AB  the  point  E  may  be  taken. 


64  FIRST   STEPS   IN   GEOMETRY 

45.    In  Fig.  89  two  parallel  lines  AB  and  CD  are  cut  by  a 
straight   line  UF.      Let  the   eight  angles    thus  formed  be 
named  by  the  letters  a,  5,  e,  d,  e,  f,  g^ 
A,  as  indicated. 

The  angles  a  and  g,  and  other  pairs 
similarly  placed  with  respect  to  AB  and 
Ci>,  are  called  exterior-interior  angles. 

The  two  pairs  of  angles  c  and  e,  d 
and/  are  called  alternate-interior  angles. 
There    are    two    very  simple    truths 
which  should  be  remembered  about  these  angles. 

1.  The  exterior-interior  ayigles  of  parallel  lines  are  equal. 

2.  The  alternate-interior  angles  of  parallel  lines  are  equal. 

EXERCISES 

1.  Name  all  the  pairs  of  exterior-interior  angles  in  Fig.  89. 

2.  Name  all  the  pairs  of  alternate-interior  angles  in  Fig.  89. 

3.  If  a  =  40°  (Fig.  89),  find  the  values  of  the  other  angles. 

4.  If  a  =  90°,  find  the  values  of  all  "the  other  angles. 

5.  If  a  =  60°,  find  the  values  of  all  the  other  angles. 

6.  Suppose  we  draw  two  parallel  lines  AB  and  CD,  and  then  draw 
a  third  line  EF  perpendicular  to  AB.  What  position  must  EF  have 
with  respect  to  CD  ? 

7.  Draw  two  parallel  lines.  Draw  a  perpendicular  to  one  of  them 
cutting  both  parallels.  Find  the  value  of  each  one  of  the  eight  angles 
thus  formed. 

8.  Draw  two  parallel  lines.  Cut  them  by  a  third  line.  Then  find 
the  values  of  the  eight  angles  formed,  using  your  protractor  as  little  as 
possible. 

9.  What  relation  exists  between  the  angles  d  and  e  in  Fig.  89  ? 
One  way  to  discover  an  answer  is  to  assume  d  equal  to  several  values, 

as  50°,  70°,  100°,  and  in  each  case  find  the  numerical  value  of  e. 

A  shorter  and  better  way  is  to  compare  d  and  e  by  applying  one  of 
the  truths  stated  above  in  italics. 


REVIKW   OF   CHAPTER   II  55 


REVIEW  EXERCISES 


1.  What  is  the  locus  of  a  point  so  moving  on  a  plane  surface  that 
it  is  always  at  the  same  distance  from  a  straight  line  on  the  surface  ? 

2.  The  radius  of  a  circle  is  5  cm.  What  is  the  locus  of  a  point  so 
moving  in  the  plane  of  the  circle  that  it  is  always  3  cm  distant  from  the 
circumference  of  the  circle  ? 

3.  Describe  a  semicircle,  draw  its  diameter,  and  join  any  three  points 
in  the  arc  to  the  ends  of  the  diameter.  Measure  the  angles,  formed 
at  the  three  points.     Do  your  results  indicate  any  general  truth  ? 

4.  A  ship  sailing  due  northeast  shifts  its  course  to  the  left  through 
one  and  a  half  right  angles.    In  what  direction  is  the  ship  now  sailing? 

5.  A  man  walks  2  miles,  then  turns  to  his  right  through  a  right 
angle  and  walks  3  miles  ;  then  turns  to  his  left  through  a  right  angle 
and  walks  1  mile.  Draw  a  plan  and  find  his  distance  from  his  starting 
point.     Take  on  the  paper  1  in.  to  represent  1  mile. 

6.  A  man  walks  6  miles,  turns  to  his  left  through  an  angle  of  45° 
and  walks  2  miles,  then  turns  through  90°  to  his  left  and  walks  2  miles. 
How  far  in  a  straight  line  is  he  now  from  his  starting  point  ?  Take  on 
the  paper  1  in.  to  represent  1  mile. 

7.  Find  the  sum  of  28°  39',  37°  48'  35",  and  78°  9'  55". 

8.  Find  the  difference  between  48°  5'  and  37°  27'. 

9.  What  is  the  supplement  of  5  times  17°  21'  ? 

10.  What  is  the  complement  of  3  times  18°  42'? 

11.  The  angular  magnitude  about  a  point  is  divided  into  seven  equal 
angles.     Find  the  value  of  each  one  correct  to  the  nearest  second. 

12.  The  circumference  of  a  circle  is  divided  into  seven  equal  parts. 
Find  the  number  of  degrees,  minutes,  and  seconds  in  each  part. 

13.  Examine  and  describe  the  following  objects  as  regards  symmetry  : 

(1)  a  horse; 

(2)  an  ordinary  door; 

(3)  a  wheel  with  a  dozen  spokes ; 

(4)  a  man  and  his  image  in  a  mirror. 

14.  Draw  a  bi-symmetric  figure  bounded  by  straight  lines. 

15.  Draw  a  bi-symmetric  figure  bounded  by  curved  lines. 

16.  Draw  a  central-symmetric  figure  bounded  by  straight  lines. 

17.  Draw  a  central-symmetric  figure  bounded  by  curved  lines. 


CHAPTER  III 

TRIANGLES  AND  QUADRILATERALS 

46.  A  plane  figure  bounded  by  three  straight  lines  is  called 
a  triangle.  The  bounding  lines  are  called  the  sides  ;  the  points 
of  intersection  of  the  sides  are  called  the  vertices ;  the  sum  of 
the  lengths  of  the  sides  is  called  the  perimeter;  and  the  angles 
included  by  the  sides  are  called  the  angles. 

A  triangle  is  called  equilateral,  if  all  the  sides  are  equal ; 
isosceles,  if  two  sides  are  equal;  scalene,  if  no  two  sides  are 
equal ;  acute,  if  all  the  angles  are  acute ;  right,  if  one  angle  is 
a  right  angle ;  obtuse,  if  one  angle  is  an  obtuse  angle. 


Isosceles 
and 
right 


Scalene  and  right 


Firx.  90. 


The  equal  sides  of  an  isosceles  triangle  are  called  the  legs, 
and  the  third  side  is  called  the  base. 

The  two  sides  that  include  the  right  angle  in  a  right 
triangle  are  called  the  legs ;  the  third  side,  the  hypotenuse. 

56 


TRIANGLES   AND   QUADRILATERALS 


57 


47.  The  dimensions  of  a  triangle  are  the  base  and  the 
altitude.  Any  side  of  a  triangle  may  be  taken  as  the  base. 
The  altitude  is  the  perpendicular  from  the  vertex  to  the  base  or 
to  the  base  produced. 


B        A 


Fig.  91. 


Fig.  92. 


In  Fig.  91  the  base  is  AB  and  the  altitude  is  CD. 
In  the  obtuse  triangle  ABC  (Fig.  92)  two  of  the  altitudes, 
AF  and  B£J,  lie  outside  of  the  triangle. 


Fig.  93. 


Describe  the  kinds  of  triangles  in  Fig.  93. 


58 


FIRST    STEPS   m    GEOMETRY 


48.  A  plane  figure  bounded  by  four  straight  lines  is  called 
a  quadrilateral. 

The  terms  sides,  vertices,  perimeter,  and  angles  have  the 
same  meaning  in  quadrilaterals  as  in  triangles. 

A  diagonal  of  a  quadrilateral  is  a  straight  line  that  joins  two 
opposite  vertices. 

A  quadrilateral  is  called  a  parallelogram,  if  the  opposite  sides 
are  parallel;  a  trapezoid,  if  two,  and  only  two,  sides  are  parallel; 
a  trapezium,  if  no  two  sides  are  parallel. 

A  parallelogram  is  called  a  rectangle,  if  the  angles  are  all 
right  angles ;  a  rhomboid,  if  the  angles  are  not  right  angles. 


\       Isosceles  trapezoid        /                    Rhomboid 

Rhombus 

• 

Right 
trapezoid 

Square 

Rectangle 

Fig.  94. 


A  rectangle  is  called  a  square,  if  the  four  sides  are  equal. 

A  rhomboid  is  called  a  rhombus,  if  the  four  sides  are  equal. 

The  parallel  sides  of  a  trapezoid  are  called  the  bases,  and 
the  non-parallel  sides  are  called  the  legs. 

A  trapezoid  is  called  an  isosceles  trapezoid,  if  the  legs  are 
equal ;  a  right  trapezoid,  if  one  leg  is  perpendicular  to  the  bases. 

49.  Any  side  of  a  parallelogram  may  be  taken  as  the  base. 
The  altitude  of  a  parallelogram  is  the  perpendicular  distance 
from  the  base  to  the  opposite  side. 


TRIANGLES   AND    QUADRILATERALS 


59 


The  line  DE  is  the  altitude  of  ABCD  (Fig.  95). 

The  line  DB  is  a  diagonal  of  the  parallelogram  ABCD. 


Fig.  95. 


FlQ.  96. 


The  altitude  of  a  trapezoid  is  the  perpendicular  distance 
between  the  bases. 

Thus,  DE  is  the  altitude  of  ABCD  (Fig.  96). 
BD  is  a  diagonal  of  the  trapezoid  ABCD. 


1    ■ 

•  / 

1 

s 

/ 

1                              1 
^             / 

5 

v./ 

'V 

A 

\  " 

/       ^^ 

N 

/ 

13 

\ 

6 

19 

\ 

\ 

j    1 

18 

\ 

17 

y\ 

V 

1  12 

9 

8 

7 

Fig.  97. 


Describe  the  different  kinds  of  quadrilaferals  in  Fig.  97. 


60  FIRST    STEPS  IN   GEOMETRY 

50.  Sum  of  the  Angles  of  a  Triangle.  When  a  straight 
line  is  made  to  rotate  about  one  of  its  points  until  its  direc- 
tion is  opposite  to  that  which  it  has  at  the  start,  the  amount 
of  rotation  is  180°.  With  this  fact  in  mind  let  us  find  the 
sum  of  the  angles  of  a  triangle. 

Take  any  triangle  ABC  (Fig.  98),  and  let  a,  h,  c  denote 
the  values  of  the  angles  that  have  as  vertices  the  points  A, 
B,  C\  respectively.  Place  a  straight 
line  DE  upon  the  side  A  C  and  regard 
it  as  having  the  direction  from  D  to 
E.  Turn  BJE  clockwise  (as  shown  by 
the  curved  arrows),  first  about  A, 
through  the  angle  a,  until  it  covers 
the  side  AB;  then  about  i?,  through 
■PiG^Qs.  ^^®  angle  5,  until  it  covers  the  side 

BC;  and  then  about  (7,  through  the 
angle  c,  until  it  covers  again  the  side  AC.  Evidently  JJE 
will  now  have  its  direction  opposite  to  that  which  it  had  at 
first.  Hence,  it  must  have  revolved  through  exactly  180°. 
It  is  obvious  that  the  whole  amount  of  rotation  reckoned  in 
degrees  is  equal  to  a -\-  b  +  c.  Therefore,  a  -\- h  -^  c  =  180°. 
We  thus  arrive  at  the  very  useful  truth:  The  sum  of  the 
angles  of  a  triangle  is  equal  to  two  right  angles^  or  180°. 

EXERCISES 

1.  How  many  right  angles  can  a  triangle  have  ?  How  many  obtuse 
angles  ?     How  many  acute  angles  ?     Give  your  reason  in  each  case. 

2.  One  angle  of  a  triangle  is  30°.  What  is  the  sum  of  the 
other  two? 

3.  Two  angles  of  a  triangle  are  25°  and  75°.     Find  the  third  angle. 

4.  If  one  acute  angle  of  a  right  triangle  is  40°,  what  is  the  value  of 
the  other  acute  angle? 


TRIANGLES   AND   QUADRILATERALS 


61 


5.  What  is  meant  by  the  statement  that  one  acute  angle  of  a  right 
triangle  is  the  complement  of  the  other  ? 

6.  If  the  acute  angles  of  a  right  triangle  are  equal,  what  is  the  value 
of  each  angle  ? 

7.  One  angle  of  a  right  triangle  is  20°.    If  all  the  sides  of  the  triangle 
are  extended,  find  the  values  of  the  new  angles  thus  formed. 

8.  Two  angles  of  a  triangle  are  each  40°.     Find  the  third  angle. 

9.  The  values  of  two  angles  of  a  triangle  are  46°  37'  and  54°  17'. 
Find  the  value  of  the  third  angle. 

10.    Give  the  name  of  a  right  triangle  if  each  acute  angle  is  45°. 

The  fact  that  the  sum  of  the  angles  of  a  triangle  is  just 
180°  may  be  shown  to  the  eye  as  follows: 

Draw  a  triangle  ABC  (Fig.  99),   and  the   altitude    CD. 
Cut  away  the  triangle  from  the  paper  (j 

and  fold  the  corners  over  the  dotted 
lines  shown  in  the  figure  until  they 
come  together  at  the  point  D.  It 
will  be  seen  that  the  three  angles  of 
the  triangle  make  at  D  a  straight  ^ 
angle,  or  two  right  angles.  fig.  99. 

51.    Sum  of   the  Angles  of   a   Quadrilateral.     Take  any 
quadrilateral  ABCD  (Fig.  100).     Divide  it  into  two  triangles 
cr  by  drawing  the  diagonal  A  C.     The 

sum  of  the  angles  of  these  two  tri- 
angles is  evidently  just  the  same 
as  the  sum  of  the  angles  of  the 
quadrilateral.  Since  the  sum  of 
'b  the  angles  of  one  triangle  is  180°, 
the  sum  of  the  angles  of  two  tri- 
angles is  2  X  180°,  or  360°. 

The  sum  of  the  angles  of  a  quadrilateral  is  360°,  or  four 
right  angles.     . 


Fig.  100. 


62  FIRST    STEPS   IN   GEOMETRY 


EXERCISES 

1.  If  the  values  of  three  angles  of  a  quadrilateral  are  40°,  70°,  and 
120°,  respectively,  what  is  the  value  of  the  fourth  angle? 

2.  If  the  angles  of  a  quadrilateral  are  all  equal,  what  is  the  value  of 
each  angle  ?     What  kind  of  a  quadrilateral  is  it  ? 

3.  If  three  of  the  angles  of  a  quadrilateral  are  right  angles,  what 
is  the  value  of  the  fourth  angle  ? 

4.  If  two  angles  of  a  quadrilateral  are  supplementary,  what  is  true 
of  the  other  two  angles  ?         • 

6.  Two  angles  of  a  quadrilateral  are  supplementary,  and  another 
angle  is  equal  to  66°.     What  is  the  value  of  the  fourth  angle  ? 

52.  Some  truths  relating  to  triangles  and  quadrilaterals 
will  now  be  stated  without  proofs.  The  pupil  should  know 
them  and  apply  them  before  he  comes  to  a  logical  study  of 
the  subject.  Each  truth,  or  theorem  as  it  is  called,  should  be 
illustrated  by  a  figure  properly  drawn  and  lettered. 


THEOREMS 

1.  The  sum  of  the  acute  angles  of  a  right  triangle  is  equal 
to  90°. 

2.  The  middle  point  of  the  hypotenuse  of  a  right  triangle  is 
equidistant  from  the  three  vertices  of  the  triangle. 

3.  The  angles  opposite  the  equal  sides  of  an  isosceles  triangle 
are  equal. 

4.  If  two  angles  of  a  triangle  are  equals  the  sides  opposite 
the  equal  angles  are  equal  and  the  triangle  is  isosceles. 

5.  An  isosceles   triangle    is   divided    into    two    equal  right 
triangles  hy  a  perpendicular  from  the  vertex  to  the  base. 

6.  The  three  angles  of  an  equilateral  triangle  are  equal. 

7.  The  three  altitudes  of  an  equilateral  triangle  are  equal. 


TRIANGLES   AND   QUADRILATERALS  63 

8.  An  equilateral  triangle  is  divided  into  two  equal  right 
triangles  hy  each  of  its  altitudes. 

9.  Two  triangles  are  equal: 

(1)  If  two  angles  and  the  included  side  of  one  are  equals 
respectively^  to  two  angles  and  the  iricluded  side  of  the  other. 

(2)  If  two  sides  and  the  included  angle  of  one  are  equals 
respectively^  to  two  sides  and  the  included  angle  of  the  other. 

(3)  If  the  three  sides  of  one  are  equals  respectively^  to  the 
three  sides  of  the  other. 

(4)  If  the  triangles  are  right  triangles  and  have  a  leg  and  the 
hypotenuse  of  one  equals  respectively.,  to  a  leg  and  the  hypotenuse 
of  the  other. 

10.  The  opposite  sides  of  a  parallelogram  are  equals  and  the 
opposite  angles  are  equal. 

11.  A    diagonal    divides    a   parallelogram   into    two    equal 
triangles. 

12.  The  diagonals  of  a  parallelogram  bisect  each  other. 

13.  The  diagonals  of  a  rectangle  are  equal. 

14.  The  diagonals  of  a  rhombus  bisect  each  other  at  right 
angles. 

15.  The  diagonals  of  a  square  are  equal  and  also  perpen- 
dicular to  each  other. 

16.  The  two  angles  at  each  base  of  an  isosceles  trapezoid  are 
equal. 

EXERCISES 

1.  How  can  a  right  triangle  be  divided  into  two  isosceles  triangles  ? 

2.  If  an  angle  at  the  base  of  an  isosceles  triangle  is  40°,  find  the 
values  of  the  other  angles. 

3.  If  the  angle  at  the  vertex  of  an  isosceles  triangle  is  50°,  find  the 
values  of  the  other  angles. 

4.  What   is   the   value  of  each  acute    angle  of  an  isosceles  right 
triangle  ? 


64  FIRST   STEPS   IN   GEOMETRY 

5.  What  is  the  vahie  of  each  angle  of  an  equilateral  triangle  ? 

6.  If  we  divide  an  equilateral  triangle  into  two  equal  right  triangles 
by  drawing  one  of  the  altitudes,  what  are  the  values  of  the  acute  angles 
of  these  right  triangles  ?  Compare  the  length  of  the  shorter  leg  of  one 
of  these  triangles  with  the  length  of  the  hypotenuse. 

7.  An  angle  A  of  a  parallelograni  ABCD  is  40°.  Find  the  values 
of  the  other  angles. 

8.  Draw  the  diagonals  of  a  rectangle,  and  describe  the  figures  into 
which  the  rectangle  is  divided. 

9.  Draw  the  diagonals  of  a  rhombus,  and  describe  the  figures  into 
which  the  rhombus  is  divided. 

10.  Draw  the  diagonals  of  a  square,  and  describe  the  figures  into 
which  the  square  is  divided. 

11.  Are  two  triangles  equal,  if  the  three  angles  of  one  are  equal, 
respectively,  to  the  three  angles  of  the  other  ?     Illustrate  by  figures. 

12.  Draw  an  acute  triangle  and  its  three  altitudes.  Do  the  altitudes 
intersect  in  a  common  point? 

13.  Draw  an  obtuse  triangle  and  its  three  altitudes.  Do  the  altitudes 
when  produced  intersect  in  a  common  point  ? 

53.  Fundamental  Problems.  The  following  problems  are 
to  be  solved  with  ruler  and  compasses.  Lengths  given  in 
centimeters  or  inches  may  be  laid  off  with  the  graduated  ruler, 
but  in  other  cases  the  dividers  should  be  used. 

54.  Problem  i .  To  construct  a  straight  line  equal  to  a  given 
straight  line. 

Let  AB  (Fig.  101)  be 

the  given  straight  line. 

Draw   a   straight    line 

K U     CD  longer  than  AB, 

With  C  as  centre  and 

radius  equal  to  AB,  cut 
Fig. 101.  '  ^ 

CD  at  E. 

Then   CE  is  the  line  required. 


B 


TRIANGLES   AND   QUADRILATERALS 


65 


55.    Problem  2.     To  bisect  a  given  straight  line. 
Let  AB  (Fig.  102)  be  the  given  straight  line. 
With  A  and  B  as  centres  and  with  equal  radii  greater  than 
half  of  AB^  describe  arcs  intersecting  at  the  points  C  and  D. 
Draw  CD  intersecting  AB  at  M, 
Then  M  is  the  middle  point  of  AB. 


VjO 


M 


I 


Fig.  102. 


u4\ 


<?^-^. 


/  !        1 


B         4- L. 


i        / 

^   \    / 


^^^B 


Fig.  103. 


56.  Problem  3*  To  find  a  'point  which  shall  he  at  given 
distances  from  two  given  points. 

Let  A  and  B  (Fig.  103)  be  the  given  points,  m  and  n  the 
given  distances. 

With  A  as  centre  and  m  as  radius,  describe  an  arc. 

With  B  as  centre  and  n  as  radius,  describe  another  arc 
intersecting  the  first  arc  at  the  points   C  and  D. 

The  points  C  and  D  both  satisfy  the  given  condition. 


EXERCISES 

1.  Make  a  straight  line  twice  as  long  as  uiB  in  Fig.  102,  bisect  it, 
and  test  the  accuracy  of  your  work. 

2.  Draw  a  triangle,  bisect  its  sides,  and  join  each  point  of  bisection  to 
the  opposite  vertex  of  the  triangle.     Do  these  lines  intersect  in  a  point? 

3.  Under  what  condition  has  Problem  3  no  solution  ?     Under  what 
condition  has  the  problem  only  one  solution  ? 


Q6  FIRST   STEPS   IN   GEOMETRY 

57.  Problem  4.  To  draw  a  perpendicular  from  a  given 
point  to  a  given  straight  line. 

Let  AB  (Fig.  104)  be  the  given  straight  line  and  C  the 
given  point. 

With  C  as  centre  and  a  suitable  radius,  cut  AB  Sit  If  and  K. 

H  and  K  are  equidistant  from  C. 

Find  by  Problem  3  another  point  0,  equidistant  from  H 
and  K.     Draw  CO  and  produce  it  to  meet  AB  in  M. 

Then  COM  is  the  perpendicular  required. 

1.  Draw  a  triangle,  and  then  draw  its  three  altitudes.  Do  the  alti- 
tudes intersect  in  a  common  point? 


■i^ 


V 

1 

1 
1 
1 

1 

y 

'       li 

Fig.  104. 

.-'K 

c/ 


/  i  \ 


^ 


i/ 

^•---, ^'B 

Fig. 105. 


58.  Problem  5.  To  erect  a  perpendicular  at  a  given  point 
of  a  given  straight  line. 

First  Method.     Proceed  as  in  Problem  4. 

Second  Method.  Let  AB  (Fig.  105)  be  the  given  straight 
line  and  B  the  given  point  of  the  line  AB. 

Take  a  point  (7,  not  in  AB\  from  C  as  centre  and  with 
CB  as  radius,  describe  an  arc  cutting  AB  at  E. 

Draw  EC  and  extend  it  to  meet  the  arc  again  at  D. 

Draw  BD.     Then  BD  is  the  perpendicular  required. 

1.  Draw  a  triangle,  and  erect  perpendiculars  at  the  middle  of  its  sides. 
Do  the  three  perpendiculars  intersect  in  a  common  point  ? 


TRIANGLES   AND   QUADRILATERALS 


67 


59.    Problem  6.     To  bisect  a  given  angle. 

Let  BAC  (Fig.  106)  be  the  given  angle. 

From  centre  A  and  with  any  convenient  radius,  describe 
an  arc  cutting  the  sides  of  the  angle  in  D  and  E^  respectively. 

From  D  and  E  as  centres  and  with  equal  radii,  describe 
arcs  intersecting  at  F,  Draw  AF.  Then  AF  is  the  bisector 
required. 

1.  Draw  a  triangle,  and  then  bisect  its  angles.  Do  the  three  bisectors 
intersect  in  a  common  point  ? 

2.  Bisect  a  right  angle,  and  bisect  one  of  its  halves.  How  many 
degrees  are  there  in  the  angle  thus  obtained? 


Fig.  106. 


Fig.  107 


60.    Problem  7.     To  trisect  a  given  right  angle. 

Let  BAC  (Fig.  107)  be  the  given  right  angle. 

From  A  as  centre  and  with  any  convenient  radius,  describe 
an  arc  cutting  AB  at  D  and  AC  at  F. 

From  D  as  centre  and  with  the  same  radius  as  before, 
describe  an  arc  cutting  the  arc  BF  at  F.     Draw  AF. 

From  F  as  centre  and  with  the  same  radius,  describe  an 
arc  cutting  the  arc  DE  at  G.     Draw  AG. 


Then  DAG  =  GAF  =  FAE 


^BAC. 


1.  What  is  the  value  in  degrees  of  the  angle  DAG'i     What  is  the 
value  in  degrees  of  the  angle  DAF? 

2.  Construct  an  angle  of  60° ;  an  angle  of  30°. 


68 


FIRST   STEPS   m   GEOMETRY 


61.  Problem  8.  At  a  given  point  in  a  given  straight  line  to 
construct  a7i  angle  equal  to  a  given  angle. 

Let  AB  (Fig.  108)  be  the  given  straight  line,  C  the  given 
point,  and  DEF  the  given  angle. 

From  E  as  centre  and  with  any  radius,  describe  an  arc 
cutting  the  sides  of  the  angle  DEF  in  G  and  H,  respectively. 

From  C  as  centre  and  with  the  same  radius  as  before, 
describe  an  arc  iJtf,  cutting  AB  at  L, 

From  L  as  centre  and  with  a  radius  equal  to  the  distance 
of  //  from  (r,  cut  the  arc  LM  at  N. 

Draw  CN. 

Then  the  angle  BCN  is  the  angle  required. 


62.  Problem  9.  To  draw  through  a  given  point  a  line 
parallel  to  a  given  straight  line. 

Let  AB  (Fig.  109)  be  the  given  straight  line  and  C  the 
given  point. 

Join  C  to  any  point  D  of  AB. 

From  D  as  centre  and  with  DC  as  radius,  describe  the 
semicircle  AECB  and  draw  BC. 

With  A  as  centre  and  a  radius  equal  to  BC^  cut  the  semicircle 
at  E^  and  draw  EC. 

Then  the  line  EC  is  tlie  parallel  required. 


TRIANGLES  AND  QUADRILATERALS       69 

63.  Problem  lo.  To  divide  a  given  straight  line  into  a 
given  number  of  equal  parts. 

Let  AB  (Fig.  110)  be  the  given  straight  line  and  three 
the  given  number  of  parts. 

Draw  AC^  making  a  conven-    ^^ G n p 

ient  acute  angle  with  AB.  ^^^^       \ 

Lay  off   on   AC  three   equal  ^^^ 

lengths,  AD^  DE^  £JF,  and  draw  ^^v^ 

FB. 

Draw  straight  lines  through 
J)  and  F  parallel  to  FB^  meeting  ^-^ j 

AB   at   the    points    G   and   If,  fig.  no. 

respectively. 

Then  AG  =  ^H  =  HB.    The  line  AB  is  said  to  be  trisected. 

1.  Draw  a  straight  line,  and  divide  it  into  five  equal  parts. 

2.  Draw  a  straight  line,  and  divide  it  into  four  equal  parts. 

64.  Problems  of  Construction.  It  now  remains  to  apply 
the  problems  just  given  to  the  construction  of  triangles  and 
quadrilaterals.  If  angles  are  given  in  degrees,  construct 
them  with  the  aid  of  a  protractor.  Also,  if  the  measurement 
of  angles  in  degrees  is  asked  for,  use  the  protractor. 

REVIEW  EXERCISES 

1.  Construct  a  right  triangle  having  for  legs  8  cm  and  4  cm.  Meas- 
ure its  acute  angles.     What  is  their  sum  ?     What  ought  the  sum  to  be  ? 

2.  Construct  a  right  triangle  having  one  leg  6  cm  long,  and  the  acute 
angle  adjacent  to  this  leg  equal  to  50°. 

3.  Construct  a  right  triangle  with  the  hypotenuse  10  cm  long,  and 
an  acute  angle  equal  to  35°. 

4.  Construct  a  right  triangle  with  one  leg  4  cm  long,  and  the  hypot- 
enuse 9  cm  long. 


70  FIRST   STEPS   m   GEOMETRY 

5.  Construct  an  isosceles  triangle,  having  given  the  base  equal  to 

5  cm,  and  an  angle  at  the  base  equal  to  75°. 

6.  Construct  an  isosceles  triangle,  having  given  one  leg  equal  to 

7  cm,  and  an  angle  at  the  base  equal  to  40°. 

7.  Construct  an  isosceles  triangle,  having  given  the  base  equal  to 

8  cm,  and  the  angle  opposite  the  base  equal  to  120°. 

Hijit.     Find  the  value  of  an  angle  at  the  base. 

8.  Construct  an  isosceles  triangle,  having  given  the  base  equal  to 

6  cm,  and  one  leg  equal  to  9  cm. 

Hint.     Erect  a  perpendicular  at  the  middle  of  the  base. 

9.  Construct  an  isosceles  right  triangle  with  legs  8  cm  long. 

10.  Construct  an  isosceles  right  triangle  with  a  hypotenuse  8  cm  long. 
Hint.     The  middle  point  of  the  hypotenuse  is  equally  distant  from 

the  three  vertices  of  the  triangle. 

11.  Construct  an  equilateral  triangle  having  a  side  6  cm  long. 

12.  Construct  an  equilateral  triangle  having  an  altitude  of  6  cm. 

13.  Construct  a  triangle  having  two  sides  8  cm  and  5  cm,  and  the 
included  angle  50°. 

14.  Construct  a  triangle  having  angles  40°  and  60°,  and  the  included 
side  7  cm  long.     What  must  the  third  angle  be  ? 

15.  Draw  with  the  ruler  a  straight  line  and  two  acute  angles.  Then 
construct  a  triangle  having  the  acute  angles  for  the  angles  and  the 
straight  line  for  the  included  side. 

16.  Construct  a  triangle  having  for  sides  6  cm,  8  cm,  and  4  cm. 

17.  Construct  a  triangle  having  for  sides  6  cm,  8  cm,  and  10  cm. 
Find  by  measurement  the  values  of  the  three  angles  of  the  triangle. 

18.  Draw  three  unequal  straight  lines.  Then  construct  a  triangle 
having  these  lines  as  sides. 

19.  Construct  a  square  having  a  side  6  cm  long. 

20.  Construct  a  square  having  a  diagonal  10  cm  long. 

21.  Draw  a  straight  line  AB.  Then  construct  a  square  having  its 
perimeter  equal  to  uiB. 

22.  Construct  a  rectangle  with  adjacent  sides  8  cm  and  5  cm. 

23.  Construct  a  rectangle  having  one  side  6  cm  long,  and  a  diagonal 
10  cm  long.     What  is  the  length  of  the  other  side? 

24.  Construct  a  rectangle  having  one  side  6  cm  long,  and  a  diagonal 
12  cm  long.     Find  the  angle  formed  by  the  diagonals. 

25.  Construct  a  rhombus,  given  one  side  6  cm,  and  one  angle  40°. 


REVIEW   OF   CHAPTER  III  71 

26.  Construct  a  rhombus,  given  one  side  10  cm,  and  one  diagonal 
8  cm. 

27.  Construct  a  rhombus  having  one  side  equal  to  one  of  the  diagonals. 
Into  what  kind  of  figures  does  this  diagonal  divide  the  rhombus? 

28.  Construct  a  rhomboid  having  its  adjacent  sides  9  cm  and  5  cm, 
and  the  included  angle  50°. 

29.  Construct  a  parallelogram  having  its  diagonals  7  cm  long  and 
12  cm  long,  and  intersecting  so  as  to  form  an  angle  of  55°. 

30.  Construct  a  parallelogram  having  a  base  8  cm  long,  and  the  alti- 
tude equal  to  G  cm.     Is  more  than  one  solution  of  thiS"  problem  possible  ? 

31.  Construct  an  isosceles  trapezoid,  given  one  base  10  cm,  one  leg 
7  cm,  and  the  angle  between  them  60°. 

32.  Construct  a  right  trapezoid,  given  the  bases  9  cm  and  5  cm,  and 
the  leg  inclined  to  the  bases  10  cm. 

33.  Construct  a  quadrilateral  having  two  adjacent  sides  each  equal 
to  8  cm,  the  other  two  sides  each  equal  to  4  cm,  and  the  angle  formed 
by  the  first  two  sides  equal  to  35°.     What  kind  of  a  quadrilateral  is  it? 

34.  Make  a  plan  of  two  vertical  posts  18  ft.  high  and  24  ft.  apart, 
with  a  horizontal  cross-bar  10  ft.  from  the  ground.  Represent  one 
foot  by  a  line  0.5  cm  long.  ■ 

35.  Make  a  plan  of  a  wall  20  ft.  high  and  64  ft.  long  with  a  door 
through  the  middle  10  ft.  high  and  16  ft.  wide.  Represent  1  ft.  by 
a  line  0.2  cm  long. 

36.  A  stone  dam  is  20  ft.  high,  34  ft.  wide  at  the  bottom,  and  4  ft. 
wide  at  the  top.  The  slant  height  on  each  side  is  the  same.  Draw  to 
scale  a  cross-section  of  the  dam,  and  find  by  measurement  the  slant 
height.     What  kind  of  a  plane  figure  is  the  cross-section? 


CHAPTER   IV 

CIRCLES  AND  REGULAR  POLYGONS 

65.  Definitions.  A  central  angle  is  an  angle  formed  by  two 
radii. 

A  sector  is  the  portion  of  a  circle  bounded  by  two  radii  and 
the  arc  contained  between  them. 

A  sector  whose  central  angle  is  a  right  angle  is  called  a 
quadrant. 

A  segment  is  the  portion  of  a  circle  bounded  by  an  arc 
and  its  chord. 

EXERCISES 

1.  Name  in  Fig.  Ill  a  central  angle,  "a  sector,  a  quadrant,  and  a 
segment. 

2.  Into  how  many  quadrants  can  a 
circle  be  divided? 

3.  Is  a  semicircle  a  sector  or  a  seg- 
ment? 

4.  Into  what  figures  is  a  circle  divided 
by  drawing  a  chord  ?  2)\ 

5.  Into    what     figures    is    a    sector 
divided   by    the    chord  of  its  arc? 

6.  Construct  a  quadrant. 

7.  Construct  a  sector  whose  angle  is 
45°. 

8.  Construct  sectors  of  30°  and  120°,  and  compare  them  as  regards 
magnitude. 

9.  Construct  a  segment  whose  chord  is  equal  to  the  radius  of  the 
circle.  What  kind  of  a  triangle  is  formed  by  drawing  radii  to  the  ends 
of  the  chord  ? 

72 


CIRCLES   AND   REGULAR   POLYGONS 


73 


66.  Radius  and  Chord.  A  sector  AOB  (Fig.  112)  is  a 
bi-symmetric  figure  with  respect  to  the  radius  OC  drawn 
perpendicular  to  the  chord  AB. 


THEOREMS 

1.  A  radius  perpendicular  to  a  chord  bisects  the  chord  and 
the  arc  subtended  by  it, 

2.  A  radius  that  bisects  a  chord  is  perpendicular  to  it. 

3.  A  perpendicular  erected  at  the  middle  point  of  a  chord 
passes  through  the  centre  of  the  circle. 

Every  theorem  contains  an  hypoth- 
esis or  statement  assumed  to  be  true, 
and  a  conclusion  or  statement  to  be 
proved  to  be  true.  Thus,  in  the 
first  of  the  theorems  just  stated  the 
hypothesis  is  that  the  radius  0(7  is 
perpendicular  to  the  chord  AB,  and 
the  conclusions  are  that  AB  =  BD, 
and  the  arc  AC  =  the  arc  BC. 


Fig, 112. 


EXERCISES 

1.  What  is  the  hypothesis  and  what  is  the  concUision  in  the  second 
of  the  theorems  above  stated  ? 

2.  What  is  the  hypothesis  and  what  is  the  conclusion  in  the  tliivd 
theorem  ? 

3.  If  in  a  circle  perpendiculars  are  erected  at  the  middle  points  of 
two  chords  not  parallel,  where  must  be  their  point  of  intersection  ? 

4.  Describe  a  circle  passing  through  three  given  points.     For  what 
position  of  the  points  is  this  problem  impossible  ? 

5.  How  many  circles  can  be  described  through  two  given  points? 

6.  What  is  the  locus  of  the  centres  of  all  circles  that  can  be  described 
through  two  given  points  ? 


74 


FIRST   STEPS   IN   GEOMETRY 


67.    Arc  and  Chord. 


Fig.  113. 


In  the  sector  AOB  (Fig.  113)  draw 
OC  perpendicular  to  AB.  Then  OC 
bisects  AB  (p.  73),  and  measures  the 
distance  of  AB  from  0  (p.  52).  Let 
the  sector  AOB  be  turned  in  the 
plane  of  the  paper  about  0  to  a 
new  position,  DOE.  Since  the  sector 
does  not  change  in  size  or  in  shape, 
the  arcs  DE  and  AB  are  equal,  the 
chords  DU  and  AB  are  equal,  and  the 
perpendiculars  OF  and  OC  are  equal. 


THEOREMS 

1.  Equal  arcs  are  subtended  by  equal  chords. 

2.  Equal  chords  subtend  equal  arcs. 

3.  Equal  chords  are  equally  distant  from  the  centre. 

EXERCISES 

1.  State  the  hypothesis  and  the  conclusion  in  each  of  the  above 
theorems. 

2.  If  angle  DOE  =  angle  A  OB,  what  is  to  be  inferred  respecting 
the  corresponding  arcs  and  chords? 

3.  Draw  an  arc,  and  then  construct  an  equal  arc. 


5.  Upon  a  given  straight  line  as  chord  construct  an  arc  of  60°. 

6.  If  the  arc  AB  (Fig.  113)  is  made  to  slide  round  the  circumfer- 
ence of  the  circle,  what  will  be  the  locus  of  the  middle  point  C  of 
its  chord  .IB? 

7.  Describe  a  circle,  and  then  draw  through  a  point  within  the  circle 
the  longest  and  the  shortest  chords  that  can  be  drawn. 

8.  Describe  a  circle,  and  then  draw  through  a  point  within  the  circle 
two  chords  equal  in  length. 

9.  Of  two  unequal  chords,  which  is  nearer  to  the  centre? 


CIRCLES    AND   REGULAR   POLYGONS 


75 


68.  Inscribed  Angles.  An  inscribed  angle  is  an  angle  whose 
vertex  is  on  the  circumference  of  a  circle,  and  whose  sides 
are  chords. 

The  angle  BAC  (Fig.  114)  is  an  inscribed  angle.  The 
chord  BC  divides  the  circle  into  two  segments,  and  the  angle 
BAC  is  said  to  be  inscribed  in  the  segment  BADC. 

THEOREMS 

1.  An  inscribed  angle  is  measured  hy  half  the  arc  inter- 
cepted between  its  sides. 

2.  An  angle  inscribed  in  a  semicircle  is  a  right  angle. 


Fig.  114 


Theorem  2  may  be  proved  as  follows : 

Inscribe  an  angle  ABC  (Fig.  115)  in  a  semicircle,  and  draw 
the  radius  OB.  The  triangles  OB  A  and  OBC  are  isosceles. 
Therefore,  the  two  angles  marked  a  are  equal,  and  the  two 
angles  marked  b  are  equal  (p.  62). 

The  sum  of  the  angles  of  the  triangle  ABC=  180°. 

In  other  words,  2  a  +  2b  =  180°.     That  is,  a  +  5  =  90°. 

But  a-{-  b  is  equal  to  the  inscribed  angle  ABC. 

Therefore,  ABC  =90°. 


76  FIRST    STEPS    IN    GEOMETKY 


EXERCISES 

1.  What  follows  from  the  first  theorem  on  page  75  as  regards  all 
angles  inscribed  in  the  same  segment?     Illustrate  by  a  figure. 

2.  Draw  a  circle  and  divide  it  into  two  unequal  segments.  Inscribe 
an  angle  in  each  segment.  Is  the  angle  inscribed  in  the  greater  seg- 
ment acute  or  obtuse  ?  Is  the  angle  inscribed  in  the  smaller  segment 
acute  or  obtuse  ?     What  reasons  for  your  answers  can  you  give  ? 

3.  Draw  a  circle  and  divide  it  into  two  unequal  segments.  Inscribe 
an  angle  in  each  segment.  'Now,  it  is  a  fact  that  the  sum  of  these  two 
angles  is  exactly  180°.  Can  you  discover  by  the  aid  of  the  first  theorem 
on  page  75  the  reason  why  this  is  true? 

4.  Construct  a  right  triangle  having  given  the  hypotenuse  =  10  cm, 
one  leg  =  6  cm. 

Draw  a  straight  line  10  cm  long,  and  upon  this  line  as  a  diameter 
describe  a  semicircle,  then  with  one  end  of  the  line  as  a  centre  and 
with  a  radius  equal  to  6  cm  in  length  describe  an  arc  cutting  the  semi- 
circumference.  The  point  of  intersection  is  the  vertex  of  the  right 
angle  of  the  triangle  required. 

5.  Construct  a  right  triangle  having  given  the  hypotenuse  =  10  cm, 
one  acute  angle  =  30°. 

6.  Construct  a  right  triangle  having  given  the  hypotenuse  =  10  cm, 
the  altitude  upon  the  hypotenuse  as  base  =  3  cm.  What  is  the  greatest 
value  that  the  altitude  could  have  in  this  problem  ? 

7.  Draw  an  isosceles  right  triangle  with  a  hypotenuse  10  cm  long. 

8.  Construct  a  right  angle  whose  sides  pass  through  two  given 
points.     Is  there  more  than  one  solution? 

9.  Two  straight  railroads  on  a  level 

plain  meet  at  a  point  A  (Fig.  116)  and 

form  an  angle  of  30°.     On  one  of  them 

there  is  a  station  B,   3   miles  from  A 

and  another  station  C,  11  miles  from  A. 

Find  a  point  D,  on  the  other  line,  such 

that   the  angle  BDC  shall  be  a  right 

angle.      Is   there  more  than  one   such 

"  Fig.  116. 

point?      Show  that  the  problem  could 

not  be  solved  if  A  C  were  7  miles.     Take  1  cm  to  represent  1  mile. 


CIRCLES   AND   REGULAR   POLYGONS 


77 


69.  Secant  and  Tangent.  A  straight  line  that  cuts  a  cir- 
cumference in  two  points  is  called  a  secant. 

A  straight  line  that  does  not  cut  a  circumference,  but  has 
one  point  in  common  with  it,  is  called  a  tangent.  The  com- 
mon point  is  called  the  point  of  contact. 

In  Fig.  117,  ABC  is  a  secant  and  ^T  is  a  tangent.  The 
point  T  is  the  point  of  contact. 
If  we  imagine  the  secant  ABC 
to  turn  about  A  till  the  points  B 
and  C  coincide,  the  secant  will 
become  a  tangent. 

A  tangent  and  a  circumfer- 
ence are  said  to  touch  each 
other  at  the  point  of  contact. 
Evidently  all  points  of  a  tangent, 
except  the  point  of  contact,  lie 
outside  the  circumference. 


THEOREMS 

1.  A  tangent  is  perpendicular  to  the  radius  drawn  to  the 
point  of  contact. 

2.  The  perpendicular    to   a   radius  at  its    extremity/  is  a 
tangent  to  the  circle. 

3.  The  perpendicular  to  a  tangent  at  the  point  of  contact 
passes  through  the  centre  of  the  circle. 


EXERCISES 

1.  State  the  hypothesis  and  the  conclusion  in  each  of  the  Theorems 
1,  2,  and  3. 

2.  Describe  a  circle,  and  then  draw  through  a  point  on  the  circum- 
ference a  tangent  to  the  circle. 

3.  If  tangents  are  drawn  through  the  ends  of  a  diameter  of  a  circle, 
what  is  their  position  with  respect  to  each  other  ? 


78  FIRST   STEPS   IN    GEOMETRY 

70.  Two  Tangents.  The  figure  FAOB  (Fig.  118)  formed 
by  two  tangents  PA^  PB^  drawn  from  an  exterior  point  P  to 
the  circle  whose  centre  is  0,  and  the  radii  OA^  OB,  drawn  to 
the  points  of  contact,  is  a  bi-symmetric  figure  with  respect 
to  the  straight  line  PO,  which  joins  P  to  the  centre. 

Hence,  PA  =  PB,  and  PO  bisects  the  angle  APB. 

THEOREMS 

1.  Two  tangents  drawn  from  an  exterior  point  to  a  circle 
are  equal. 

2.  Two  tangents  from  an  exterior  point  make  equal  angles 
with  the  straight  line  which  joins  the  exterior  point  to  the 
centre  of  the  circle. 


Fig.  118  Fig.  119. 

EXERCISES 

1.  What  are  the  values  of  the  angles  PAO  and  PBO  (Fig.  118)? 

2.  What  is  the  sum  of  the  angles  APO  and  .4 OP? 

3.  If  ^ Pi?  =  90°,  what  kind  of  a  figure  is  APBO'i 

4.  Describe  a  circle,  and  then  draw  tangents  to  the  circle  through 
a  point  P  outside  the  circumference  (Fig.   119). 

Upon  PO  as  diameter,  describe  a  circle  cutting  the  given  circle  at 
the  points  A  and  B.     Draw  the  straight  lines  PA  and  PB. 

5.  Draw  two  tangents  to  a  circle  making  an  angle  of  60°. 


CIRCLES   AND   REGULAR  POLYGONS 


79 


71.  Railroad  Curves.  In  laying  out  a  railroad  curve, 
care  must  be  taken  to  avoid  a  sudden  change  of  direction 
at  the  points  where  the  curve  begins  and  ends.  At  these 
points,  therefore,  the  straight  track  must  be  so  joined  to  the 
curve  that  it  is  a  tangent  to  the  curve.  How  the  connec- 
tion may  be  made  is  illustrated  in  Fig.  120. 


oy 


/45' 


Fig.  120. 


A  railroad  AB  runs  due  east,  and  it  is  desired  to  connect 
with  it  a  branch  line  CD  which  runs  southeast.  In  this 
case  the  total  change  of  direction  is  45°.  Starting  from  the 
point  C,  the  engineer  lays  off  equal  distances  C£J  and  CF 
along  the  lines,  taking  into  account  the  fact  that  the  greater 
these  distances  are,  the  less  rapid  will  be  the  change  in 
direction  along  the  curve.  At  the  points  B  and  F,  perpen- 
diculars are  erected  meeting  at  the  point  0.  If  now  with  O 
as  centre  and  OF  as  radius,  an  arc  FF  is  laid  out,  it  is  clear 
that  the  straight  lines  AF  and  F>F  will  be  tangents  to  this 
arc.  Thus,  a  sudden  change  of  direction  at  the  points  F 
and  F  is  avoided. 


80 


FIRST   STEPS  IN   GEOMETRY 


72.  Two  Circles.  The  relative  position  of  any  two  circles 
is  determined  by  the  distance  between  their  centres  as  com- 
pared with  the  sum  of  their  radii.  Let  d  denote  the  dis- 
tance between  their  centres,  r  and  r'  their  radii. 

There  are  six  cases  to  be  considered. 

CdBe  1.  d  greater  than  r  +  /  (Fig.  121).  In  this  case  the 
two  circles  lie  wholly  outside  each  other. 

Case  2.     d=^r-\-r'   (Fig.  122).     In  this   case  the  circles 


Fig.  121. 


Fig.  122. 


touch  each  other  externally,  and  a  tangent  common  to  both 
circles  can  be  drawn  through  the  point  of  contact. 

Case  S,     d  less  than  r  +  r\  but  greater  than  r  —  r'  (Fig. 
123).     In  this  case  the  circles  intersect  in  two  points. 


Fig.  123. 


Fig. 124. 


Fig. 125. 


Case  4'  d  =  r  —  r'  (Fig.  124).  The  circles  touch  inter- 
nally, and  a  tangent  common  to  both  circles  can  be  drawn 
through  the  point  of  contact. 


CIRCLES   AND   REGULAR   POLYGONS 


81 


Case  5.     d  less  than  r  —  r'  (Fig.  125).     In  this  case  the 
smaller  circle  lies  wholly  within  the  larger. 

Case  6.     li  d  =  0,  the  two  circles  are  concentric. 


EXERCISES 

How  many  common  tangents  can  be  drawn  to  two  circles: 

1.  When  the  circles  lie  wholly  outside  each  other? 

2.  When  the  circles  touch  each  other  externally? 

3.  When  the  circles  intersect  each  other  ? 

4.  When  the  circles  touch  each  other  internally  ? 

5.  When  one  circle  lies  wholly  within  the  other  circle  ? 

73.  Ornamentai  Curves.  By  joining  arcs  of  circles,  various 
useful  ornamental  curves  may  be  constructed.  The  following 
are  some  examples. 


Fig. 126. 


Fig.  127. 


Fig. 129. 


EXERCISES 

1.  Construct  curves  like  those  in  Figs.  126  and  127, 

2.  Construct  an  oval  like  that  in  Fig.  128. 

3.  Construct  a  spiral  like  that  in  Fig.  129. 


82 


FIRST    STEPS   IN   GEOMETRY 


74.  Polygons.  A  plane  figure  bounded  by  straight  lines 
is  called  a  polygon.  This  definition  includes  the  triangle  and 
the  quadrilateral,  but  the  term  ''  polygon  "  is  usually  applied 
to  plane  figures  having  more  than  four  sides. 

Polygons  have  special  names  according  to  the  number  of 
their  sides.  A  pentagon  has  5  sides,  a  hexagon  has  6  sides, 
an  octagon  has  8  sides,  a  decagon  has  10  sides,  a  dodecagon  has 
12  sides,  an  icosagon  has  20  sides. 

The  sides,  the  perimeter,  and  the  vertices  of  a  polygon  are 
defined  exactly  as  in  the  case  of  a  triangle. 

A  diagonal  of  a  polygon  is  a 
straight  line  joihing  two  vertices 
not  adjacent. 

By  drawing  all  the  diagonals 
possible  from  one  vertex  of  a 
polygon,  we  can  divide  the  poly- 
gon into  triangles  ;  the  number  of 
triangles  thus  obtained  is  always 
two  less  than  the  number  of  sides. 

The  sum  of  the  angles  of  the  pentagon  in  Fig.  130  is 
evidently  equal  to  the  sum  of  the  angles  of  the  triangles 
into  which  it  is  divided;   that  is,  equal  to  3  X  180°. 

The  sum  of  the  angles  of  a  polygon  is  equal  to  180°  multi- 
plied hy  the  number  of  sides  minus  two. 


Fig.  130. 


EXERCISES 

1.  Find  the  sum  of  the  angles  of  a  hexagon. 

2.  Find  the  sum  of  the  angles  of  an  octagon. 

3.  Find  the  sum  of  the  angles  of  a  decagon. 

4.  Find  the  sum  of  the  angles  of  a  dodecagon. 

5.  Draw  a  hexagon,  and  then  draw  all  the  diagonals  that  can  be 
drawn.     How  many  diagonals  are  there  ? 


CIRCLES   AND   REGULAR   POLYGONS 


83 


75.  Regular  Polygons.  A  regular  polygon  is  a  polygon  that 
has  all  its  sides  equal  and  all  its  angles  equal. 

There  exists  in  every  regular  polygon  a  point  equidistant 
from  all  the  vertices,  and  also  equidistant  from  all  the  sides. 
This  point  is  called  the  centre  of  the  polygon. 

An  angle  at  the  centre  of  a  regular  polygon  is  the  angle 
formed  by  two  straight  lines  drawn  from  the  centre  to  the 
ends  of  one  side.  The  angles  at  the  centre  of  a  regular 
polygon  -are  equal. 

Straight  lines  from  the  centre  of  a  regular  polygon  to  all 
the  vertices  divide  the  polygon  into  as  many  equal  isosceles 


E 

K        D 

H- 

/  V 

/  "^-.^ 

/  -"'''^    \ 

/       ^ 

'.  \ 

/  y^                      \ 

\ 

;A--..    / 

^\  / 

V 

A        G         B 
Fig.  132. 


triangles  as  there  are  sides ;  and  by  drawing  perpendiculars 
from  the  centre  to  the  sides,  the  polygon  is  divided  into 
twice  as  many  equal  right  triangles  as  there  are  sides.  In 
this  way  a  regular  pentagon  (Fig.  131)  is  divided  into  five 
equal  isosceles  triangles  and  ten  equal  right  triangles. 

A  regular  polygon  has  central  symmetry.  The  centre  of 
the  polygon  is  the  centre  of  symmetry.  The  symmetry  is  as 
many  fold  as  there  are  sides. 

Regular  polygons  also  have  axial  symmetry.  A  regular 
hexagon  (Fig.  132)  has  six  axes  of  symmetry. 


84  FIRST   STEPS   IN   GEOMETRY 


EXERCISES 

1.  How  can  the  angle  at  the  centre  of  a  regular  polygon  be  found? 

2.  Find  the  values  of  the  angles  at  the  centre  of  regular  polygons 
having  3,  4,  5,  6,  8,  10,  and  100  sides. 

3.  Find  the  value  of  an  angle  of  a  regular  hexagon. 

First  find  the  sum  of  all  the  angles  of  the  hexagon  (see  p.  82),  and 
then  divide  the  sum  by  6. 

4.  Find  the  value  of  an  angle  of  a  regular  octagon. 

5.  Find  the  valiie  of  an  angle  of  a  regular  decagon. 

6.  How  can  the  centre  of  a  given  regular  polygon  be  found? 
Two  methods  are  suggested  by  Figs.  131  and  132. 

76.  Inscribed  and  Circumscribed  Figures.  A  polygon  is 
said  to  be  inscribed  in  a  circle  if  its  sides  are  chords  of  the 
circle,  and  the  circle  is  said  to  be  circumscribed  about  the  polygon. 

A  polygon  is  said  to  be  circumscribed  about  a  circle  if  its 
sides  are  tangents  to  the  circle,  and  the  circle  is  said  to  be 
inscribed  in  the  polygon  (Figs.  133  and  134). 


In  the  case  of  a  regular  polygon,  the  centre  of  the  polygon 
coincides  with  the  centres  of  the  inscribed  and  the  cir- 
cumscribed circles.  The  easiest  way  to  construct  a  regular 
polygon  is  to  describe  a  circumference,  divide  it  into  as  many 
equal  arcs  as  there  are  sides  in  the  polygon,  and  draw  the 
chords  of  these  arcs. 


CIRCLES   AND   REGULAR   POLYGONS 


85 


EXERCISES 

1.    Circumscribe  a  circle  about  a  given  rectangle  (Fig.  135), 
To  solve  this  problem  a  point  must  be  found  equidistant  from  the 
four  vertices  of  the  rectangle. 

D 


Fig.  135 


2.  Inscribe  a  circle  in  a  given  rhombus  (Fig.  136). 

The  diagonals  bisect  the  angles  of  the  rhombus,  and  the  point  0 
where  the  diagonals  intersect  is  the  centre  of  the  inscribed  circle. 

3.  Draw  an  isosceles  trapezoid,  and  inscribe  a  circle  in  it. 


Fig,  137. 


Fig.  138. 


4.    Construct  Fig.  137.     Begin  by  drawing  the  large  square. 

6.    Construct  Fig.  138.     Begin  by  drawing  the  equilateral  triangle. 


86 


FIRST    STEPS    IN    GEOMETRY 


77.  Division  of  a  Circumference.  The  problem  of  con- 
structing a  regular  polygon  depends  for  its  solution  on  our 
ability  to  divide  a  circumference  into  equal  arcs.  The  theo- 
rems of  elementary  geometry  enable  us  to  make  this  division 
with  the  aid  of  ruler  and  compasses  in  three  general  cases. 

Oase  1.     Division  into  2,  4,  8,  16,  32,  etc.,  equal  arcs. 

A  circumference  is  divided  into  two  equal  arcs  by  simply 
drawing  a  diameter,  and  into  four  equal  arcs  by  drawing  two 
diameters  perpendicular  to  each  other.  If  these  arcs  are 
bisected,  the  circumference  will  be  divided  into  eight  equal 
arcs.  In  practice  we  bisect  one  of  the  four  arcs,  and  lay  off 
the  chord  of  half  the  arc  eight  times  round  the  circumference, 
as  shown  in  Fig.  139. 


Fig.  139. 


Fig.  140. 


Case  2.     Division  into  3,  6,*  12,  24,  48,  etc.,  equal  arcs. 

Draw  a  radius  OA  (Fig.  140).  Lay  off  a  chord  AB  =  OA. 
Draw  OB.  The  triangle  AOB  is  equilateral,  and  therefore 
the  angle  AOB  =  60°.  Since  60°  is  contained  in  360°  just  six 
times,  it  follows  that  the  radius  of  a  circle  carried  as  a  chord 
round  the  circumference  will  divide  it  into  six  equal  arcs. 

The  first,  third,  and  fifth  points  trisect  the  circumference. 


CIRCLES   AND   REGULAR   POLYGONS 


87 


Case  3.     Division   into  5,   10,   20,  40,   etc.,  equal  arcs. 

Draw  two  perpendicular  diam- 
eters AB  and  CD  (Fig.  141). 
Bisect  the  radius  OA  at  E^  and 
with  E  as  centre  and  the  dis- 
tance EC  as  radius,  describe  an 
arc  cutting  OB  at  F.  CF  applied 
as  a  chord  five  times  will  divide 
the  circumference  into  five  equal 
arcs,  and  OF  applied  as  a  chord 
ten  times  will  divide  it  into  ten 
equal  arcs. 

The  proof  involves  reasoning 
of  too  difficult  a  character  to  be  given  here. 


Fig.  141. 


EXERCISES 


1.  Inscribe  a  square  in  a  circle. 

2.  Inscribe  a  regular  octagon  in  a  circle. 

3.  Circumscribe  a  square  about  a  circle. 

4.  Circumscribe  a  regular  octagon  about  a  circle. 

5.  Inscribe  a  regular  hexagon  in  a  circle. 

6.  Inscribe  an  equilateral  triangle  in  a  circle. 

7.  Circumscribe  a  regular  hexagon  about  a  circle. 

8.  Circumscribe  an  equilateral  triangle  about  a  circle. 

9.  Inscribe  an  equilateral  triangle  in  a  circle,  and  circumscribe 
an  equilateral  triangle  about  the  same  circle.  Are  the  four  triangles 
thus  formed  equal  triangles? 

10.  Inscribe  a  square  in  a  circle,  and  circumscribe  a  square  about 
the  same  circle.     Draw  the  diagonals  of  the  inscribed  square. 

11.  Inscribe  a  regular  hexagon   in  a  circle,  and   circumscribe   an 
equilateral  triangle  about  the  same  circle. 

12.  Inscribe  a  regular  pentagon  in  a  circle. 

13.  Circumscribe  a  regular  decagon  about  a  cinOe. 


88 


FIRST   STEPS   IN    GEOMETRY 


78.  Stone  pavements  illustrate  the  use  of  regular  poly- 
gons. Not  every  regular  polygon  can  be  used ;  only  those 
can  be  used  whose  sides  touch  at  all  points,  for,  otherwise, 
the  surface  would  not  be  everywhere  covered. 


TTTTTTTT 

TTTTTTTV 

TTTTTTTT 
'TTTTTTT' 
TTTTTTTT 
^TTTTTTT^ 


Fig.  142. 


Fig. 143. 


Fig. 144. 


Equilateral  triangles  can  be  used  (Fig. 
142)  in  groups  of  six  about  a  point ;  for 
each  angle  is  60°,  and  six  of  these  angles 
fill  360°,  the  whole  angular  magnitude 
about  a  point.  The  arrangement  in 
squares  is  shown  in  Fig.  143.  Regular 
hexagons  may  be  arranged  in  groups  of 
three  about  a  point  (Fig.  144). 


tMTj^ 


XXXJL 


Fig.  145. 


f!g.  146. 


In  Fig.  145  regular  hexagons  and  equilateral  triangles  are 
combined. 

In  Fig.  146  regular  octagons  and  squares  are  combined. 


CIRCLES   AND   REGULAR   POLYGONS 


89 


79.  Problems  of  Construction.  We  add  a  few  more  ex- 
amples of  the  numerous  ways  in  which  straight  lines  and 
arcs  of  circles  may  be  combined  so  as  to  produce  ornamental 
figures. 

1.  Construct  a  regular  hexagon.  Then,  taking  as  centres  the  centre 
of  the  hexagon  and  its  six  vertices,  describe  circles,  each  with  a  radius 
equal  to  half  of  one  side  of  the  hexagon.  Lastly,  describe  a  circle 
which  shall  enclose  the  whole  figure  and  touch  six  of  the  smaller 
circles. 

2.  Construct  a  square.  Draw  its  diagonals.  With  the  intersec- 
tion of  the  diagonals  as  centre  and  with  a  radius  equal  to  one  fourth  of 
a  diagonal,  describe  a  circle.  With  the  vertices  of  the  square  as  centres 
and  with  the  same  radius  as  before,  describe  arcs  within  the  square  and 
bounded  by  its  sides.  Join  the  ends  of  the  four  arcs  by  straight  lines. 
In  this  way  a  regular  octagon  is  formed  which  may  be  said  to  be 
inscribed  in  the  square. 


3.  Construct  Fig.  147.  The  distances  AB,  BC  are  equal.  The 
arcs  described  with  A,  B,  etc.,  as  centres  have  equal  radii.  The  curved 
forms  produced  in  this  fashion  are  characteristic  of  Gothic  architecture. 


ATAYAYAYA 


TT 
Fig.  148. 


4.    Construct  Fig.  148.     The  distances  AB,  BC,  etc.,  are  equal  to 
twice  the  distance  between  the  two  horizontal  parallels. 


90 


FIRST   STEPS   IN    GEOMETRY 


5.    Construct  a  five-starred  figure  (Fig.  149).     Begin  by  describing 
a  circle  and  marking  the  vertices  of  an  inscribed  regular  pentagon. 


Fig.  149. 


6.  Construct  a  compass  face  with  eight  rays  (Fig.  150).  Describe 
a  circle  and  mark  the  vertices  of  an  inscribed  regular  octagon. , 

7.  Construct  the  design  shown  in  Fig.  151.  'Describe  first  the  large 
circle.  Each  arc  within  the  circle  has  the  vertex  of  the  inscribed 
regular  octagon  for  centre  and  a  side  of  the  octagon  for  radius. 


Fig.  151.  Fig.  152. 

8.   Construct  the  ornamental  design  shown  in  Fig.  152. 


CIRCLES   AND   RP^GULAR   POLYGONS  91 

80.  Length  of  a  Circumference.  The  length  of  the  cir- 
cumference of  a  circle  is  greater  than  the  perimeter  of  an 
inscribed  regular  polygon  and  less  than  the  perimeter  of  a 
circumscribed  regular  polygon.  For  example,  the  perimeter 
of  an  inscribed  regular  hexagon  is  six  times  the  length  of  the 
radius  of  the  circle  or  three  times  the  length  of  its  diameter, 
but  the  perimeter  of  the  circumscribed  square  is  four  times 
the  length  of  the  diameter.  Therefore,  the  length  of  the 
circumference  must  be  more  than  three  times,  but  less  than 
four  times,  the  length  of  the  diameter. 

To  find  very  nearly  the  relation  between  the  circumference 
of  a  circle  and  its  diameter  in  respect  to  length,  geometers 
take  advantage  of  the  fact  that  the  greater  the  number  of 
sides  of  an  inscribed  or  a  circumscribed  regular  polygon,  the 
less  the  difference  between  the-  length  of  its  perimeter  and 
the  length  of  the  circumference.  They  suppose  regular 
polygons  of  a  great  number  of  sides  inscribed  in  the  circle 
and  circumscribed  around  the  circle,  and  then  compare  the 
lengths  of  their  perimeters  with  the  diameter  of  the  circle. 

In  this  way  it  can  be  proved  that  the  length  of  a  circum- 
ference is  more  than  3.141592  times',  but  less  than  3.141593 
times,  the  length  of  the  diameter.  If  tlien  we  multiply  the 
diameter  by  3.14159,  we  shall  obtain  very  nearly  the  length 
of  the  circumference.  In  fact,  the  error  made  by  using  this 
value  to  compute  the  circumference  of  a  circle  whose 
diameter  is  1  mile  is  less  than  1  inch. 

The  approximate  value  usually  taken  for  the  ratio  of  the 
circumference  of  a  circle  to  the  diameter  is  3.1416,  or  3^. 
The  true  value  of  the  ratio  cannot  be  expressed  exactly  in 
figures  but  is  represented  by  the  Greek  letter  ir  (Pi). 

It  follows  that  the  circumference  of  a  circle  is  equal  to  ir 
times  the  diameter. 


92  FIRST   STEPS    IN    GEOMETRY 


REVIEW   EXERCISES 

With  English  units  take  it  =  ^^\  with  metric  units  take  it  =  3.1416. 

1.  Find  the  circumference  of  a  circle  if  the  diameter  is  35  ft. 

2.  Find  the  circumference  of  a  circle  if  the  radius  is  14  in. 

3.  Find  the  circumference  of  a  circle  if  the  radius  is  8  cm. 

4.  Find  the  diameter  of  a  circle  if  the  circumference  is  132  ft. 

5.  Find  the  radius  of  a  circle  if  the  circumference  is  1  m. 

6.  If  the  driving  wheel  of  a  locomotive  has  a  radius  of  3  ft., 
how  many  revolutions  will  it  make  in  going  236  miles? 

7.  What  is  the  diameter  of  a  circular  pond  if  a  man  take  840  paces 
of  2^  ft.  each  in  walking  around  the  pond  ? 

8.  What  should  be  the  diameter  of  a  round  dining  table  for  eight 
persons,  allowing  3  ft.  for  each  person? 

9.  The  earth  turns  on  its  axis  once  every  24  hours.  Taking  its 
diameter  as  8000  miles,  how  far  does  a  point  on  the  equator  move  in 
1  second? 

10.  The  circumferences  of  two  concentric  circles  are  16|  ft.  and 
18  ft.     Find  the  width  of  the  ring  between  them. 

11.  The  radius  of  a  circle  is  7  ft.  What  is  the  length  of  an  arc 
of  30°,  or  yijj  of  the  circumference  ? 

12.  If  the  radius  of  a  circle  is  14  in.,  find  the  length  of  an  arc 
of  45°. 

13.  The  diameter  of  a  circle  is  5  ft.  10  in.  and  the  angle  formed  by 
two  radii  is  150°.     Find  the  length  of  the  included  arc. 

14,.  The  radius  of  a  circle  is  28  in.  Find  the  value  of  the  central 
angle  that  will  subtend  an  arc  11  in.  long. 

Solution.    Circumference  =  -2^2_  x  2  x  28  in.  =  176  in.,  and  j^V  =  tV 
Therefore,  the  central  angle  =  j\  of  360°  =  22°  30'. 

15.  The  radius  of  a  circle  is  8  ft.  2  in.  and  the  length  of  an  arc  is 
6  ft.  5  in.     Find  the  central  angle  that  subtends  this  arc. 

16.  A  man,  walking  4  miles  per  hour,  walks  for  15  minutes  on  the 
circumference  of  a  circle,  the  radius  of  which  is  1  mile.  Find  the 
length  of  the  arc  passed  over  and  the  number  of  degrees  in  it. 

17.  Plow  many  degrees,  etc.,  are  there  in  an  arc  equal  in  length  to 
the  radius  of  the  circle  ? 


CHAPTER   V 


AREAS 


81.    Equivalent  figures  are  figures  that  have  the  same  size. 

Equal  figures  agree  both  in  size  and  in  shape,  but  two 
equivalent  figures  may  have  very  different  shapes. 

For  example,  the  square  and  the  triangle  in  Fig.  153  are 
quite  unlike  in  shape,  yet  they  have  been  so  drawn  that 
they  enclose  equal  amounts  of 
surface.  Therefore,  they  are 
equivalent  figures. 

But  what  right  have  we  to 
assert  that  the  square  and  the 
triangle  represented  in  Fig.  153 
have  exactly  the  same  size? 
This  question  suggests  the 
more  general  question:  How 
can  we  compare  as  regards  size 
two  figures  that  differ  in  shape  ? 

To  answer  this  question  for  plane  figures  is  the  object  of 
this  chapter. 

In  other  words,  we  are  now  to  learn  how  plane  figures  are 
measured.  In  one  respect  the  method  resembles  that  employed 
for  measuring  lines.  We  first  choose  a  suitable  unit^  and 
then  proceed  to  find  the  number  of  times  this  unit  is  con- 
tained in  the  surface  to  be  measured. 

The  number  of  units  of  surface  thus  found  is  called  the 
area  of  the  surface. 


Fig.  153. 


94 


FIRST   STEPS   IN    GEOMETRY 


82.    Units  of  Surface.     The  units  of  surface  generally  used 
are  squares  whose  sides  are  equal  to  the  units  of  length. 

The  English  units  of  surface  are : 
The  square  inch  (sq.  in.). 
The  square  foot  (sq.  ft.)     =  144  sq.  in. 
The  square  yard  (sq.  yd.)  =      9  sq.  ft. 
The  square  rod  (sq.  rd.)     =    30^  sq.  yd. 
The  acre  (A.)  =  160  sq.  rd. 

The  square  mile  (sq.  mi.)  =  640  A. 

The  most  important  metric  units  of  surface  are : 
The  square  millimeter  (qmm). 
The  square  centimeter  (qcm)  =100  qmm. 
The  square  decimeter  (qdm)   =100  qcm. 
The  square  meter  (qm)  =100  qdm. 

The  square  kilometer  (qkm)    =  1,000,000  qm. 
(Approximately,  1  qcm  =  J  sq.  in. ;  1  qm  =  li  sq.  yd.) 


EXERCISES 

.    1.    One    centimeter  =  10    millimeters.     Why   does    it    follow    that 
1  square  centimeter  contains  100  square  millimeters? 

Fig.  154  should  enable  you  to  answer 
this  question.  Each  side  of  the  square 
A  BCD  is  to  be  considered  as  representing 
1  cm,  and  is  divided  into  ten  equal  parts. 
Perpendiculars  to  the  sides  are  drawn  at 
all  the  points  of  division. 

2.  Explain  by  a  diagram  why  a  square 
foot  contains  144  sq.  in. 

3.  Why  is  a  square  yard  equal  to  9 
sq.  ft.? 

4.  Show  that  an  acre  contains  43,560 
sq.  ft.  Fig.  154. 


^    ........         D 


AREAS 


95 


D 


I j -r 


Fig.  155. 


83.  Area  of  a  Square.  Let  one  side  of  a  square  (Fig.  155) 
be  4  in.  long.  Divide  two  adjacent  sides  into  inches,  and 
erect  perpendiculars  at  all  the  points  of 
division.  The  square  is  thus  divided 
into  4  X  4  or  16  smaller  squares,  each 
equal  to  1  sq.  in.  Therefore,  the  area  of 
the  given  square  is  equal  to  16  sq.  in. 

Suppose  that  one  side  of  the  square 
is  4.25  in.  Proceed  as  before,  only 
suppose  each  side  divided  into  425  equal 
parts,  each  j^-^  of  an  inch  long.  Evidently  we  shall  obtain 
425  X  425  or  180,625  smaller  squares.  But  it  will  take 
100  X  100  or  10,000  of  these  little  squares  to  make  1  sq.  in. 
Therefore,  the  area  of  the  square  in  square  inches  will  be 
'TWoV"  ^^  18.0625.  This  result  is  obtained  by  multiplying 
the  number  4.25  by  itself. 

In  general  the  area  of  a  square  is  found  by  multiplying 
the  number  of  units  of  length  in  one  side  of  the  square  by 
itself;  in  other  words,  by  squaring  the  number  of  units  of 
length  in  one  side. 

This  truth,  for  the  sake  of  brevity,  may  be  stated : 

Area  of  a  square  ==  square  of  one  side. 


EXERCISES 

Find  the  area  of  a  square  if : 

1.  One  side  =    9  in.  4.  One  side  =  15  cm. 

2.  One  side  =  14  ft.  5.  One  side  =  25  m. 

3.  One  side  =    2^  in.  6.  One  side  =  6.42  cm. 

7.  What  will  it  cost  at  20  cents  a  square  foot  to  paint  a  square 
floor,  the  side  of  which  is  18^  ft.  long? 

8.  Find  the  area  in  acres  and  in  square  feet  of  a  square  field,  each 
side  of  which  is  178  yd.  long. 


96 


FIRST   STEPS   IN    GEOMETRY 


84.  Area  of  a  Rectangle.  Let  ABCD  (Fig.  156)  be  a 
rectangle  in  which  AB  =  7  in.  and  AD  =  4  in.  Then  by  the 
D  c  same  method  as  in  the  case  of  a 

square  it  can  be  shown  that  the 
area  of  the  rectangle  is  equal  to 
4  X  7,  or  28,  sq.  in. 

In  every  case  the  area  of  a  rect- 
angle is  found  by  multiplying  the 
number  of  units  of  length  in  one 
side  by  the   number  of  units  of 
length  in  the  adjacent  side.     The  two  adjacent  sides  of  a 
rectangle  are  usually  called  the  length  and  the  breadth. 
We  may  express  the  above  rule  by  the  formula, 

Area  of  a  rectafigle  =  length  X  breadth. 


-1 \ 1 —J H y. 


Fig.  156. 


EXERCISES 

Find  the  area  of  a  rectangle,  having  given : 

1.  Length  12  in.,  breadth  9  in. 

2.  Length  15^  in.,  breadth  6  in. 

3.  Length  3^  in.,  breadth  2^  in. 

4.  Length  200  ft.,  breadth  60  ft. 

5.  Length  200  ft.,  breadth  10  yd. 

6.  Length  16  cm,  breadth  12  cm. 

7.  Length  40  cm,  breadth  2  dm. 

Find  the  breadth  of  a  rectangle,  having  given  : 

8.  Area  100  sq.  in.,  length  20  in. 

9.  Area  288  sq.  in.,  length  24  in. 

10.  Area  22^  sq.  in.,  length  9  in. 

11.  Area  180  sq.  ft.,  length  6  yd. 

12.  Determine  as  accurately  as  you  can  the  number  of  square  inches 
of  surface  on  the  cover  of  this  book. 

13.  Find  the  area  of  a  football  field,  if  the  length  is  110  yd.  and 
the  width  is  53^  yd. 


AREAS  97 

14.  How  many  bricks  9  in.  by  4  in.  are  required  to  cover  a  floor 
34  ft.  long  and  17  ft.  wide? 

Note.  The  expression  "  9  in.  by  4  in."  means  "  9  in.  long  and  4  in. 
wide."     Therefore,  the  surface  covered  by  each  brick  is  36  sq.  in. 

15.  How  many  square  feet  of  plank  are  required  for  a  floor  20  ft. 
long  and  14  ft.  wide?  How  many  boards,  each  14  ft.  long  and  6  in. 
wide,  are  needed?  How  many  yards  of  carpet  28  in.  wide  are  required 
for  covering  the  floor  ? 

16.  A  lawn  measuring  144  yd.  by  98  yd.  is  to  be  covered  with 
turf.  Each  turf  is  cut  18  in.  long  and  15  in.  wide.  The  price  to  be 
paid  is  75  cents  a  dozen  laid  down.  Find  the  cost  of  turfing  the 
lawn. 

17.  What  length  must  be  taken  on  a  plank  18  in.  wide  to  obtain  just 
4  sq.  ft.  of  plank  ? 

18.  What  must  be  the  width  of  a  field  half  a  mile  long  to  contain 
just  100  acres  ? 

19.  How  is  the  area  of  a  rectangle  changed  by  doubling  the  length  ? 
By  doubling  the  breadth?  By  doubling  both  the  length  and  the 
breadth  ? 

20.  From  a  rectangular  lot  528  ft.  by  240  ft.  there  are  sold  four 
square  lots,  each  having  a  side  60  ft.  long.     How  much  land  remains  ? 

21.  A  square  and  a  rectangle  have  equal  perimeters,  60  m.  The 
rectangle  is  twke  as  long  as  it  is  wide.  Which  figure  has  the  greater 
area,  and  what  is  the  difference  between  the  areas  ? 

22.  The  sides  of  a  rectangle  are  4  in.  and  9  in.  Find  the  side  of  an 
equivalent  square. 

23.  The  sides  of  a  rectangle  are  4  in.  and  10  in.  Find  the  side  of 
an  equivalent  square,  correct  to  two  decimal  places. 

24.  A  man  has  a  rectangular  field  400  ft.  by  240  ft.  If  he  makes 
streets  60  ft.  wide  directly  through  the  field  parallel  to  the  sides,  so  as 
to  divide  the  field  into  four  equal  rectangles,  and  divides  each  rect- 
angle into  two  equal  house  lots,  how  much  land  will  there  be  in  each 
house  lot?  Draw  a  plan  of  the  field  divided  as  described,  taking  1  cm 
to  represent  25  ft.      How  many  square  feet  does  1  qcm  represent? 

25.  Find  how  many  square  feet  of  boarding  will  be  required  to 
make  a  fence  40  yd.  long  and  4  boards  high,  if  the  boards  average  10  in, 
wide. 


98 


FIRST    STEPS   IN    GEOMETRY 


Fig,  157. 


85.    Area  of  a  Parallelogram.     Take  any  parallelogram, 
ABCD  (Fig.  157),  draw  AF  and  BE  perpendicular  to  AB^  and 

;py  ry  EC    6^^^^^  ^^  ^^  ^'     ^^^  ^^i^  w^y 

we  form  a  rectangle  ABJEF, 
having  the  same  base  and  the 
same  altitude  as  the  parallelo- 
gram. 

The  triangles  AFI),  BEC  are 
right  triangles  having  equal  legs  AF"  and  BE^  and  equal 
hypotenuses  AD  and  BC.  Therefore,  the  triangles  are  equal 
(p.  63).  If  the  triangle  AFD  is  taken  from  the  whole  figure 
ABCF^  there  is  left  the  parallelogram ;  if  the  equal  triangle 
BEC  is  taken  from  the  whole  figure,  there  is  left  the  rectangle. 
Hence,  the  parallelogram  and  the  rectangle  are  equivalent 
figures.  The  area  of  the  rectangle  is  equal  to  AB  X  BE 
(p.  96).  Therefore,  the  area  of  the  parallelogram  is  equal  to 
AB  X  BFJ.     In  general. 

Area  of  a  parallelogram  =  base  X  altitude. 


EXERCISES 

Find  the  area  of  a  parallelogram,  having  given : 

1.  Base  16  in.,  altitude  5  in. 

2.  Base  2^  in.,  altitude  14  in. 

3.  Base  8.2  cm,  altitude  6.7  cm. 

4.  Name  in  Fig.  1 58  three  parallelograms  and  one  rectangle.   Compare 
their  bases.    Compare  their  alti- 
tudes.     Compare    their   areas. 
What  is  true  of  parallelograms 


that  have  equal  bases  and  equal 
tiltitudes  ? 

5.  The  bases  of  two  paral- 
lelograms are  equal.  The  alti- 
tude of  one  is  four  times  that  of  the  other.     Compare  their  areas. 


AREAS 


99 


86.  Area  of  a  Triangle.  A  diagonal  divides  a  parallelo- 
gram into  two  equal  triangles  (p.  63).  Hence,  any  triangle 
ABC  (Fig.  159)  may  be  regarded  as 
half  of  a  parallelogram  ABCE  having 
the  same  base  AB  and  the  same  alti- 
tude as  the  triangle. 

The  area  of  a  parallelogram  =  base 
X  altitude  (p.  98).     Hence, 


Fig,  159. 


Area  of  a  triangle  =  ^  base  X  altitude. 


EXERCISES 

Find  the  area  of  a  triangle,  having  given  : 

1.  Base  11  in.,  altitude  10  in. 

2.  Base  2|-  in.,  altitude  1  in. 

3.  Base  20  ft.  6  in.,  altitude  10  ft.  4  in. 

4.  Base  10.4  cm,  altitude  8.6  cm. 

5.  Examine  the  triangles  in  Fig.  160.  Have  they  the  same  base? 
Have  they  equal  altitudes  ?  Have  they  equal  areas  ?  What  is  true  of 
triangles  that  have  equal  bases  and  equal  altitudes  ? 


6.  In  a  triangle  ABC  (Fig.  161),  a  straight  line  is  drawn  irom  the 
vertex  C  to  the  middle  point  D  of  the  base  AB.  Compare  the  areas  of 
the  triangles  ADC  and  BDC. 

7.  How  can  a  triangle  be  divided  into  four  equal  triangles? 

8.  Can  you  explain  how  a  triangle  may  be  divided  into  two  triangles 
such  that  one  has  double  the  area  of  the  other  ? 


100  FIRST   STEPS   IN   GEOMETRY 

87.  Area  of  a  Rhombus.  The  area  of  a  rhombus  may 
be  found  by  multiplying  its  base  by  its  altitude  just  as  in 
the  case  of  any  other  parallelogram.  But  the  diagonals 
of   a   rhombus    bisect   each   other  at  right  angles  (p.  63) ; 

and  this  fact  gives  us  another 
way  of  finding  the  area  of  a 
rhombus. 

Let  ACDE  (Fig.  162)  be  a 
rhombus.  Draw  the  diagonals 
AD^  CE.  AD  divides  the  rhom- 
bus into  two  equal  isosceles  trian- 
gles, A  CD  and  AED,  each  having 
as  base  AD  and  as  altitude  half  of  CE.  Therefore,  the  area 
of  each  triangle  =  \  AD  ^^  \  CE  =  \  AD  X  CE.  Hence,  the 
area  of  the  rhombus  =  2  X  \AD  ><  CE  =  \AD  X  CE.     Hence, 

Area  of  a  rhombus  =  half  the  product  of  the  diagonals. 

EXERCISES 

Find  the  area  of  a  rhombus,  having  given : 

1.  The  diagonals  6  in.  and  8  in. 

2.  The  diagonals  9|  in.  and  6;^  in. 

3.  The  diagonals  330  ft.  and  132  ft. 

4.  It  is  desired  to  lay  out  a  park  in  the  shape  of  a  rhombus  so  that 
it  shall  contain  10  acres,  and  the  longer  diagonal  shall  be  2420  ft.  long. 
Find  the  shorter  diagonal,  and  draw  a  plan  of  the  park,  taking  1  cm  to 
represent  200  ft. 

5.  Make  a  square  by  pinning  together  at  the  ends  four  narrow  strips 
of  cardboard.  Then  turn  the  sides  about  the  pins  so  as  to  make  the 
angles  oblique.  What  kind  of  a  figure  is  thus  formed?  If  a  square 
and  a  rhombus  have  equal  sides,  which  has  the  greater  area  ? 

6.  The  legs  of  a  right  triangle  are  6  in.  and  8  in.     Find  the  area. 

7.  Show  that  the  area  of  a  right  triangle  is  equal  to  half  the  product 
of  the  legs. 


AREAS 


101 


Fig.  163. 


88.    Area  of  a  Trapezoid.     A  trapezoid  ABCD  (Fig.  163)  is 

divided  into  two  triangles  ABD^ 
BBC   by  drawing    the   diagonal 

BD.  The   base   of  the  triangle 
ABD  is  AB^  and  the  altitude  is 

BE.  The  base  of  the  triangle 
BDC  is  i)C,  and  the  altitude  is 

BF.  which  is  equal  to  BE, 

Area  of  triangle  ABB  =  ^  AB  X  BE. 

Area  of  triangle  BBC  =  ^  BC  X  BF  =  ^  BC  X  BE. 

Adding,  we  have  area  of  trapezoid  =  i-  (AB  -\-  BC)  X  BE, 
a  result  which  we  may  state  as  follows : 

Area  of  a  trapezoid  =  ^  (sum  of  bases)  X  altitude. 
EXERCISES 


Find  the  area  of  a  trapezoid,  having  given : 

1.  Bases  10  in.  and  8  in.,  altitude  6  in. 

2.  Bases  20  cm  and  30  cm,  altitude  5  cm. 

3.  Bases  37  ft.  and  25  ft.,  altitude  19  ft. 

4.  Find  in  square  miles  the  area  of  the  trapezoid  ABCD  (Fig.  163), 
iiAB  =  ^  mile,  CD  =  ^  mile,  DE  =  \  mile. 

5.  The  four  lateral  faces  of  the  pedestal  of  a  statue  are  equal  trape- 
zoids (Fig.  164).  The  bases  of  each  trapezoid  are  20  ft.  and  15  ft.  in 
length.  The  altitude  of  each  trape- 
zoid is  equal  to  8  ft.  What  is  the 
entire  lateral  surface? 

6.  A  playground  has  the  shape 
of  an  isosceles  trapezoid.  The  bases 
are  600  ft.  and  200  ft.  in  length,  and 
the  larger  base  makes  angles  of  60° 
with  the  legs.     Construct  a  plan  of 

the  playground,  taking  1  cm  to  represent  50  ft.,  and  find  the  area  of 
the  playground  in  square  feet. 


Fio.  164. 


FIRST    STEPS   IN   GEOMETRY 


89.  Area  of  a  Regular  Polygon.  The  straight  lines  drawn 
from  the  centre  of  a  regular  polygon  (Fig.  165)  to  the 
vertices  divide  the  polygon  into  as  many 
equal  isosceles  triangles  as  the  polygon 
has  sides.  The  base  of  each  triangle  is  a 
side  of  the  polygon,  and  its  altitude  is 
equal  to  the  radius  of  the  inscribed  circle. 
If  we  find  the  area  of  one  of  these  tri- 
angles and  multiply  it  by  the  number  of 
sides  of  the  polygon,  we  evidently  obtain 
the  area  of  the  polygon. 

Approximate  values  of  the  radius  of  the  inscribed  circle  for 
several  regular  polygons  are  given  in  the  following  table  : 


No.  of  sides.  Radius. 

3  .     .  .  .  0.288  X  one  side 

4  .     .  .  .  0.500  X  one  side 

5  .     .  .  .  0.688  X  one  side 

6  .     .  .  .  0.866  X  one  side 


No.  of  sides.  Radius. 

7  .     .     .  .  1.038  X  one  side 

8  .  .  .  .  1.207  X  one  side 
10  ...  .  1.539  X  one  side 
12     .     .     .  .  1.866  X  one  side 


EXERCISES 

Find  the  area  of : 

1.  An  equilateral  triangle,  if  one  side  is  5  in. 

2.  A  regular  pentagon,  if  one  side  is  6  in. 

3.  A  regular  octagon,  if  one  side  is  15  ft. 

4.  A  regular  decagon,  if  one  side  is  19  ft. 

5.  A  park  has  the  shape  of  a  regular  hexagon.     Each  side  is  1000  ft. 
long.     What  is  the  assessed  valuation  at  8  cents  per  square  foot  ? 

6.  Find  the  total  area  covered  by  the  shaded  triangles  in  Fig.  142, 
p.  88,  if  a  side  of  each  triangle  is  2  ft.  long. 

7.  Find  the  total  area  covered  by  the  shaded  squares  in  Fig.  143, 
p.  88,  if  a  side  of  each  square  is  2  ft.  long. 

8.  One  side  of  a  regular  hexagon  is  1  ft.  long.     Find  the  side  of  a 
square  equivalent  to  the  hexagon. 


AREAS 


103 


90.  Area  of  any  Polygon.  The  area  of  any  polygon  may 
be  found  by  dividing  the  polygon  into  simpler  figures  (triangles, 
rectangles,  trapezoids,  etc.),  as  illustrated  by  the  following 
exercises,  then  computing  the  area  of  each  part,  and  adding 
the  results. 

EXERCISES 

1.  Find  the  area  of  the  polygon  ABCD  (Fig.  166),  having  given : 
AC  400  yd. ;  BE  120  yd. ;  DF  80  yd. 


Fig.  166. 


Fig.  167. 


2.  Find  the  area  of  the  polygon  ABODE  (Fig.  167),  having  given  : 
BE  108  yd. ;  EC  96  yd. ;  ^F  49  yd. ;  HD  35  yd. ;  CG  67  yd. 

3.  Find  the  area  of  the  polygon  ABCDE  (Fig.  168),  having  given  : 
AC  550  yd. ;  BG  160  yd.  ;  AF  160  yd. ;  AH  500  yd. ;  EF  160  yd. ; 
DH  90  yd. 

E 
E 


4.  Find  the  area  of  the  polygon  ABCDE F  (Fig.  169),  having  given  : 
AC  340  yd.;  BK  115  yd.;  FG  50  yd.;  EG  54  yd.;  FH  283  yd.; 
DH  60  yd. ;  AF  80  yd. ;  CF  330  yd. ;  the  angle  A  FC  =  90°. 

What  other  mode  of  division  might  be  used  for  this  polygon  ? 


104 


FIRST   STEPS   IN   GEOMETRY 


91.  Theorem  of  Pythagoras.  The  following  very  useful 
truth  was  discovered  more  than  2000  years  ago  by  the  Greek 
philosopher  Pythagoras  : 

The  square  of  the  hypotenuse  of  a  right  triangle  is  equal  to 
the  sum  of  the  squares  of  the  two  legs. 

A  special  case  of  the  theorem  is  illustrated  in  Fig.  170. 
The  sides  of  the  right  triangle  contain  3,  4,  and  5  units  of 
length,  respectively.  The  squares  constructed  upon  the  sides 
contain  9,  16,  and  25  units  of  surface,  respectively.  Since 
25  =  9  4- 16,  the  theorem  evidently  holds  true  in  this  case. 


Fig.  170 


Consider  the  case  of  a  right  triangle  ABC  having  equal 
legs  AB  and  BC  (Fig.  171).  The  acute  angles  at  A  and  C 
are  each  45°.  If  we  construct  upon  the  legs  the  squares 
ABDE  and  BCFG,  and  draw  DG,  AD,  and  CG,  we  obtain  a 
figure  ACFGDE  divided  into  six  equal  right  triangles,  each 
equal  to  the  triangle  ABC.  The  square  constructed  upon 
AC  contains  four  of  these  triangles,  and  each  of  the  squares 
constructed  upon  the  legs  contains  two  of  these  triangles. 
Therefore,  the  theorem  holds  true  in  this  case. 

The  theorem  can  be  proved  true  for  any  right  triangle. 


AREAS  105 


EXERCISES 


1.  If  the  legs  of  a  right  triangle  are  6  cm  and  8  cm  long,  find  the 
hypotenuse. 

2.  If  the  legs  of  a  right  triangle  are  8  cm  and  15  cm  long,  find  the 
hypotenuse. 

3.  The  hypotenuse  of  a  right  triangle  is  20  cm  long,  and  one  leg  is 
16  cm  long.     What  is  the  length  of  the  other  leg? 

4.  Find  the  length  of  one  leg  of  a  right  triangle  if  the  lengths  of  the 
other  leg  and  the  hypotenuse  are  5  cm  and  13  cm. 

5.  A  ladder  34  ft.  long  just  reaches  a  window  when  placed  with  its 
foot  16  ft.  from  the  side  of  the  house.  How  high  is  the  window  above 
the  ground? 

6.  How  long  must  a  ladder  be  to  reach  the  top  of  a  wall  24  ft.  high, 
if  the  foot  of  the  ladder  is  10  ft.  from  the  bottom  of  the  wall? 

7.  The  longest  side  of  a  meadow  in  the  shape  of  a  right  triangle 
cannot  be  measured  directly  on  account  of  the  swampy  nature  of  the 
ground.  What  is  its  length,  if  the  other  sides  are  960  ft.  and  720  ft, 
long? 

8.  Two  vessels  start  at  the  same  time  from  the  same  place.  One 
sails  due  north  at  the  rate  of  6  miles  an  hour,  and  the  other  sails  due 
east  at  the  rate  of  8  miles  an  hour.  How  far  apart  are  they  at  the  end 
of  6  hours  ? 

9.  Find  the  area  of  a  right  triangle  if  the  length  of  the  hypotenuse 
is  25  in.  and  the  length  of  one  leg  is  15  in. 

10.  Find  to  two  decimal  places  the  h;y'potenuse  of  a  right  triangle, 
each  leg  of  which  is  20  cm  long. 

11.  Find  to  two  decimal  places  the  legs  of  an  isosceles  right  triangle, 
If  the  hypotenuse  is  20  cm  long. 

12.  Find  the  area  of  an  isosceles  triangle  if  one  leg  is  10  ft.  long  and 
the  base  is  12  ft.  long. 

13.  The  side  of  an  equilateral  triangle  is  20  ft.     Find  the  altitude 
and  the  area  of  the  triangle. 

14.  The  side  of  a  square  is  20  ft.     Find  the  length  of  the  diagonal. 

15.  The  diagonal  of  a  square  is  20  ft.     Find  the  length  of  one  side. 

16.  How  far  apart  are  the  opposite  corners  of  a  floor  that  is  24  ft, 
long  and  18  ft.  wide? 


106  FIRST    STEPS   IN   GEOMETRY 

92.  Area  of  a  Circle.  Consider  a  regular  polygon  circum- 
scribed about  a  circle  (Fig.  172),  and  divided  into  equal 
isosceles  triangles. 

The  area  of  each  triangle  is  equal  to  the  product  of  half 
of  one  side  of  the  polygon  and  the  radius 
of  the  circle.  It  follows  that  the  area  of 
the  polygon  is  equal  to  the  product  of 
half  its  perimeter  and  the  radius  of  the 
circle.  Now  the  area  of  the  polygon  is 
greater  than  the  area  of  the  circle.  Also, 
the  more  sides  there  are  to  the  polygon 
the  more  nearly  its  area  approaches  the 
area  of  the  circle.  Starting  from  these  obvious  facts,  it  can 
be  proved  that  the  area  of  the  circle  is  equal  to  the  product  of 
half  the  circumference  and  the  radius. 

Approximately  (p.  91)  the  circumference  =  ^^-  X  diameter. 
Hence,  half  the  circumference  =  ^-f-  X  radius. 

Therefore,  the  area  of  a  circle  =  ^^-  X  square  of  radius^  or 
if  instead  of  ^^-^  we  take  the  decimal  value  3.1416, 

Area  of  a  circle  =  3.1416  X  square  of  radius. 

REVIEW  EXERCISES 

Find  the  area  of  a  circle,  having  given  : 

1.  Radius  7  in.  5.  Diameter  49  ft. 

2.  Radius  14  ft.  6.  Diameter  3^  ft. 

3.  Radius  20  cm.  7.  Diameter  16  cm. 

4.  Radius  1  m.  8.  Diameter  100  in. 

9.    The  diameter  of  a  park  in  the  shape  of  a  circle  is  half  a  mile. 
Find  the  number  of  acres  in  the  park. 

10.  A  horse  is  tied  to  a  stake  on  a  grass  plot  by  a  rope  42  ft.  long. 
Over  how  many  square  feet  of  grass  can  he  feed  ? 

11.  The  radii  of  two  concentric  circles  are  14  ft.  and  21  ft.  Find 
the  area  of  the  ring  bounded  by  the  circumferences. 


REVIEW  OF   CHAPTER  V  107 

12.  A  circular  fort  56  ft.  in  diameter  is  surrounded  by  a  moat  14  ft. 
wide,  full  of  water.     Find  the  area  of  the  moat. 

13.  There  are  two  circular  lots  of  land.  The  radius  of  one  lot  is  just 
half  that  of  the  other  lot.  If  the  first  lot  is  worth  $100,  what  is  the 
second  lot  worth  ? 

14.  How  many  circles,  each  having  a  radius  of  3  in.,  will  be  required 
to  obtain  an  area  as  large  as  that  of  a  single  circle  whose  radius  is 
12  in.? 

15.  Find  the  area  of  the  largest  circle  that  can  be  cut  from  a  square 
piece  of  wood,  each  side  of  which  is  2  ft.  4  in.  long. 

16.  The  circumference  of  a  circle  is  equal  to  the  perimeter  of  a 
square.  Which  figure  has  the  greater  area?  Show  by  an  example  that 
your  answer  is  correct. 

17.  The  area  of  a  circle  is  100  sq.  ft.     Find  the  radius. 

22 
Solution.      100  =  y  X  (radius)^  ; 

7  X  100 
(radius)2  =  — -- —  ft.  =  31.8181  ft. ; 


radius  =  V31.8181  ft.  =  5.64  ft.,  nearly. 

18.  What  must  be  the  diameter  of  a  circular  field  that  the  area  may 
be  6  acres? 

19.  Find  the  radius  of  a  circle,  if  the  area  is  44  sq.  in. 

20.  Find  the  radius  of  a  circle  equivalent  to  a  square,  each  side  of 
which  is  11  ft.  long. 

21.  The  radius  of  a  circle  is  4  in.  Find  the  radius  of  a  circle  whose 
area  is  twice  as  large  as  the  area  of  the  given  circle ;  also  the  radius  of 
a  circle  whose  area  is  four  times  as  large  as  the  area  of  the  given  circle. 

22.  Find  the  side  of  a  square  equivalent  to  a  circle,  the  diameter  of 
which  is  84  ft.  long. 

23.  Find  the  diameter  of  a  circle  equivalent  to  a  square,  one  side  of 
which  is  7  ft.  long. 

24.  Find  the  number  of  acres  of  land  enclosed  by  a  circular  race 
track  1  mile  long. 

25.  The  diameter  of  a  circular  reservoir  is  64  yd.  Around  the 
reservoir  a  walk  1  yd.  wide  is  constructed,  and  outside  the  walk  a  fence 
is  built,  find  the  length  of  the  fence  and  the  area  of  the  walk. 


108  FIRST    STEPS   IN    GEOMETRY 

26.  Find  the  area  of  a  circular  sector,  if  the  radius  of  the  circle  is 
14  in.,  and  the  angle  of  the  sector  is  45°. 

Since  the  circumference  contains  360°,  it  is  clear  that  the  sector  of 
45°  is  -^^jj,  or  ^  of  the  area  of  the  entire  circle  (Fig.  173). 
Find  the  areas  of  the  following  sectors : 

27.  Radius  7  in.,  angle  30°. 

28.  Radius  35  in.,  angle  120°. 

29.  Diameter  28  ft.,  angle  60°. 

30.  The  radius  of  a  circle  is  42  in.  Find 
the  area  of  a  "sector  whose  arc  is  22  in.  long. 

The  area  of  the  whole  circle  is  half  the 
circumference  x  the  radius,  and  the  area  of 
any  sector  is  half  its  arc  x  the  radius. 

31 .  Find  the  area  of  a  sector  whose  radius 
is  14  in.  long,  and  whose  arc  is  11  in.  long. 

What  is  the  value  of  the  angle  of  this  sector  ? 

32.  What  part  of  the  entire  circle  is  a  sector,  the  chord  of  whose 
arc  is  equal  in  length  to  the  radius  of  the  circle  ?  What  is  the  value  of 
the  angle  of  this  sector  ? 

33.  Find  the  area  of  the  segment  A  CB  (Fig.  173),  if  the  radius  of 
the  circle  is  7  in.,  and  the  angle  A  OB  is  60°. 

Evidently  segment  A  CB  =  sector  A  OB  —  triangle  A  OB. 

In  this  case  the  triangle  A  OB  is  equilateral,  and  each  side  is  equal 
to  the  radius  of  the  circle.  The  altitude  OD  of  the  triangle  bisects  the 
chord  AB,  and  its  length  can  be  found  by  applying  the  Theorem  of 
Pythagoras  (p.  104). 

34.  The  radius  of  a  circle  is  equal  to  56  ft.,  and  the  central  angle 
corresponding  to  a  segment  of  the  circle  is  90°.  Find  the  area  of  the 
segment. 

35.  The  diameter  of  the  cross-section  of  a  wooden  log  is  3  ft.  Find 
the  area  of  the  cross-section  of  the  largest  square  beam  that  can  be 
sawed  from  the  log. 


CHAPTER    VI 


SIMILAR  FIGURES 


93.  Numerical  Measures.  A  magnitude  is  measured  by 
finding  the  number  of  times  it  contains  another  magnitude  of 
the  same  kind  taken  as  a  unit  of  measure.  The  number  of 
times  the  unit  is  contained  in  the  magnitude  is  called  the 
numerical  measure  of  the  magnitude  with  reference  to  the  unit. 

If  MJSr(Fig.  174)  is  taken  as  the  unit  of  length,  and  if  AB, 
BC\  etc.,  are  each  equal  to  3£2V,  then  the  numerical  measure 
of  AB  with  reference  to  JfiV  is  1,  that  oi  AC  is  2,  etc. 

M\ \]sr 

A\ ; ; \ '— \ 1 1 

B  C  D  E  F  G  H 

Fig.  174. 

94.  Ratio.  The  ratio  of  two  quantities  of  the  same  kind 
is  the  quotient  of  the  numerical  measure  of  one  quantity- 
divided  by  the  numerical  measure  of  the  other,  when  both 
quantities  are  expressed  in  terms  of  the  same  unit. 

For  example  (Fig.  174),  the  ratio  of  AD  to  AG  is  f  or  i, 
the  ratio  oi  AG  to  AD  is  |  or  |,  the  ratio  of  ^C  to  AD  is  §. 

A  ratio  is  often  expressed  by  placing  a  colon  between  the 
two  numbers  or  terms  of  the  ratio.  Thus,  the  ratio  of  AD  to 
^6^  is  written  3 : 6,  and  read  "  as  3  is  to  6,"  or  "  the  ratio  of 
3  to  6." 

109 


110  FIRST   STEPS   m   GEOMETRY 

Find  the  ratio  of  two  straight  lines  AB,  CD  (Fig.  175). 

A\ 1 1 1 \B 

F 

C\ 1 1 1 ID 

E 

Fig. 175. 

First  Method,  Measure  each  line  to  the  nearest  milli- 
meter with  a  graduated  ruler,  and  divide  one  result  by  the 
other. 

Second  Method.  Apply  by  means  of  dividers  AB  to  CD 
as  many  times  as  possible.  In  this  case  AB  can  be  applied 
twice,  and  there  is  a  remainder  ED.  Now  apply  this  remain- 
der ED  to  AB  as  many  times  as  possible ;  in  this  case  three 
times  with  the  remainder  FB.  Apply  this  remainder  FB 
to  the  remainder  ED  ;   in  this  case  exactly  twice.     Then 

ED  =  2FBi 

AB  =  ^ED-{-  FB=  IFB; 

CD  =  2AB  +  ED  =  UFB  +  2FB  =  16FB. 

Therefore,         777-  == 


CD      16  FB     16' 
OT  AB:  CD  =  1  :  16. 

Here  FB  is  a  common  measure  for  the  lines  AB  and  CD. 

If  two  lines  have  no  common  measure,  they  are  said  to  be 
incommensurable.  The  side  and  the  diagonal  of  a  square 
are  incommensurable  lines. 

EXERCISE 

1.  What  is  the  ratio  (Fig.  174)  oi  AB  to  AC '^  oi  AC  to  AB? 
oi  AC  to  AD?  oi  AC  to  AG?  oi  AD  to  AH?  oi  AH  to  AF? 


SIMILAR   FIGURES  111 

95.  Proportion.  Four  quantities  are  said  to  be  proportional, 
or  to  form  a  proportion,  if  the  ratio  of  the  first  to  the  second 
is  equal  to  the  ratio  of  the  third  to  the  fourth. 

The  four  quantities  are  called  the  terms  of  the  proportion ; 
the  first  and  the  fourth  are  called  the  extremes ;  the  second 
and  the  third  are  called  the  means. 

Thus,  the  numbers  2,  5,  4,  10  are  proportional  numbers  ; 
the  proportion  is  written  2  :  5  =  4  :  10,  and  is  read  '^  2  is  to  5 
as  4  is  to  10  " ;  2  and  10  are  the  extremes,  5  and  4  the  means. 

The  lines  AB,  CD  (Fig.  176)  have  the  ratio  3:5;  and  the 
lines  EF,  GR  (Fig.  177)  have  the  same  ratio.  The  four  lines 
AB,  CD,  EF,  GH  form  the  proportion  AB:  CD  =  EF:  GH. 


A 

B 

D 

E 

F 

C 

G 

h 

Fig.  176. 

Fig.  177. 

In  every  proportion  the  product  of  the  extremes  is  equal  to 
the  product  of  the  means. 

Thus,  in  the  proportion  2  : 5  =  4  :  10,  the  product  of  the 
extremes  is  20,  and  the  product  of  the  means  is  20. 


EXERCISES 

i.  Express  in  lowest  terms  the  ratio  of  6  in.  to  24  in.,  4  ft.  to  6  in., 
15°  to  90°. 

2.  If  the  first  three  terms  of  a  proportion  are  36,  9,  and  28,  find  the 
fourth  term. 

9  X  28 

Let  X  denote  the  fom'th  term.     Then  x  =  ^^ ^  =  7. 

36 

Find  the  missing  term  in  the  following  proportions  : 

3.  6  :  12  =  20  :  X.  6.    a: :  40  =  17  :  68. 

4.  4:24=a::90.  7.    2.5  : 7.5  =  0.5  :  a:. 

5.  l:x  =  9:36.  8.    3^  :  28^  =  1 :  a:. 


112 


FIRST    STEPS   IN   GEOMETRY 


96.  In  certain  cases  it  can  be  proved  that  the  ratio  of  two 
areas  is  equal  to  the  ratio  of  two  lines.  By  the  use  of  pro- 
portions of  this  kind  the  solution  of  numerical  questions 
respecting  areas  is  often  simplified. 


EXERCISES 


1.  "Two  triangles  ADC,  BDC  (Fig.  178)  have  the  same  altitude  CE. 
Their  bases  are  13  ft.  long  and  20  ft.  long. 

Compare  their  areas.  We  know  (p.  99)  that 

triangle  ADC  =  lSxiCE, 

and  triangle  BDC  =  20  x  ^CE. 

triangle  ^Z)C      IZxhCE      13 

Therefore  2 =  f =  — 

'  triangle  J5Z>C      20x^CE      20 

That  is, 

triangle  ADC :  triangle  BDC  =  13  :  20. 

2.  If  the   area   of  the  triangle  ADC 

(Fig.  178)  is  208  sq.  ft.,  find  the  area  of  the  triangle  BCD. 

3.  Compare  (Fig."  178)  the  areas  of  the  triangles  ABC  and  ADC; 
ialso  the  areas  of  the  triangles  ABC  and  BDC. 

4.  The  bases  of  two  triangles  that  have  equal  altitudes  are  5  ft.  and 
15  ft.,  respectively.  If  the  smaller  triangle  contains  100  sq.  ft.,  what  is 
the  area  of  the  larger  triangle  ? 

5.  What  is  always  the  ratio  of  the  areas  of  two  triangles  that  have 
equal  altitudes  but  unequal  bases  ? 

6.  A  triangular  field  ABC  contains  just  8  acres.  The  owner  wishes 
to  sell  2  acres  of  it,  and  to  run  the  division  line  straight  from  C  to  the 
proper  point  D  in  AB.  Show  how  to  locate  the  point  D  without  making 
any  measurements  in  the  field. 

7.  Two  triangles  have  equal  bases.  The  altitude  of  one  is  10  ft. ; 
that  of  the  other  is  60  ft.  Show  by  proceeding  as  in  Ex.  1  that  one 
triangle  is  six  times  as  large  as  the  other. 

8.  The  altitudes  of  two  triangles  that  have  equal  bases  are  7  ft.  and 
63  ft.  The  area  of  the  smaller  triangle  is  84  sq.  ft.  Find  the  area  of 
the  larger  triangle,  and  find  the  common  base. 


SIMILAR   FIGURES  113 

9.    What  is  always  the  ratio  of  the  areas  of  two  triangles  that  have 
equal  bases  but  unequal  altitudes  ? 

10.  Draw  any  triangle  and  then  construct  in  the  simplest  way 
another  triangle  which  shall  be  three  times  as  large. 

11.  Two  parallelograms  have  equal  altitudes,  but  the  base  of  one  is 
ten  times  as  long  as  that  of  the  other.     Compare  their  areas. 

12.  A  triangle  and  a  parallelogram  have  equal  bases  and  equal  alti- 
tudes.    What  is  the  ratio  of  their  areas  ? 

13.  A  triangle  and  a  parallelogram  have  equal  bases  and  equal  areas. 
What  is  the  ratio  of  their  altitudes  ? 

14.  A  triangle  and  a  parallelogram  have  equal  bases.  The  altitude  of 
the  triangle  is  four  times  that  of  the  parallelogram.  Compare  their 
areas. 

15.  It  is  desired  to  run  a  division  line  AE  (Fig.  179)  straight  across 
a  rectangular  field  A  BCD,  so  that  the  triangle  ADE  cut  off  shall  be 
equal  in  area  to  i  of  the  entire  field. 
How  would  you  determine  the  proper 
position  of  the  point  ^?  What  is  the 
ratio  of  DE  to  DC?  What  is  the  ratio 
of  DE  to  EC 'I 

Locate  the  point  E  (Fig.  179)  in  order 
that:  Fig.  179. 

16.  Triangle  ADE  shall  be  one  fourth  the  rectangle  A  BCD. 

17.  Triangle  ADE  :  rectangle  ABCD  =  1 :  12. 

18.  Triangle  ADE  :  rectangle  ABCD  =  3:7. 

19.  The  radii  of  two  circles  are  1  ft.  and  3  ft.,  respectively.  Find 
the  ratio  of  the  circumferences  and  the  ratio  of  the  areas. 

20.  If  the  radius  of  a  circle  is  doubled,  what  change  takes  place  in 
the  circumference?     W^hat  change  in  the  avea? 

21.  The  diameters  of  two  circles  are  as  1  to  5.  What  is  the  ratio  of 
their  circumferences  ?     What  is  the  ratio  of  their  areas  ? 

22.  The  diameters  of  two  circular  flower  beds  are  as  3  to  4.  The  area 
of  the  smaller  bed  is  90  sq.  ft.     AVhat  is  the  area  of  the  other? 

23.  The  radii  of  three  circular  water  tanks  are  6  ft.,  9  ft.,  and  15 
ft.,  respectively.  If  a  mason  charges  %o  for  cementing  the  bottom  of 
the  smallest  tank,  what  ought  he  to  charge  for  cementing  the  bottom  of 
each  of  the  other  tanks  ? 


A 

B 

\ 

D 

E  . 

C 

114 


FIRST   STEPS   IN   GEOMETRY 


97.  Draw  an  angle  XAY  (Fig.  180),  lay  off  on  one  side 
any  equal  lengths  AB,  BC,  CD,  etc.  Through  the  points  B, 
C,   D,   etc.,   draw  any  parallel  lines,   cutting  the  side  AY 

at  the  points  E,  F,  G,  etc. 
The  more  carefully  you 
draw  these  parallels  the  more 
nearly  you  will  find  by  using 
dividers  that  the  lengths  AE, 
EF,  FG,  etc.,  are  also  equal. 
In  this  way  you  will  doubt- 
less become  convinced  of  the 
truth  of  the  following  theo- 
rem, although  the  proof  is  not  given  here. 

If  parallel  lines  intercept  equal  lengths  on  one  side  of  an 
angle',  they  also  intercept  equal  lengths  on  the  other  side  of  the 
angle. 

This  theorem  is  one  of  the  most  useful  truths  of  geometry. 
Upon  it  is  based  the  division  of  a  straight  line  into  equal 
parts  (p.  69),  and  it  leads  directly  to  the  idea  of  proportional 
lines,  and  to  the  solution  of  various  problems  of  great 
practical  importance. 

It  is  easy  to  find  proportional  lines  in  Fig.  180. 

Since  AB  =  BC  =  CD, 

we  have  AB:BD  =  1:2. 

And  since  AF  =  FF  =  FG, 

we  have  AF:FG  =  1:2. 

Therefore,  AB :  BD  =  AF :  FG. 

Again,  AB:AD  =  1:S,  and  AE:AG  =  1:S. 

Therefore,  AB :  AD  =  AF :  AG. 

Show  by  reasoning  in  a  similar  manner  that 
AC :  CD  =  AF  :  FG. 


SIMILAR    FIGURES 


115 


Fig,  181. 


98.  A  straight  line  drawn  parallel  to  one  side  of  a  triangle, 
cutting  the  other  two  sides,  divides  those  sides  proportionally/ . 

This  follows  from  the  theorem  on  page  114. 

For  example,  let   BE  (Fig.   181)   be  ^ 

drawn  parallel  to  the  side  BC  oi  the  tri- 
angle ABC. 

Then      AB:BB  =  AE :  EC, 

AB:AB  =  AEiAC, 

and  AB:BB  =  ACiEC. 

Upon  this  truth  depends  the  solution  of  the  following 
problems. 

EXERCISES 

1.  Divide  a  given  straight  line  AB  (Fig.  182)  into  two  parts  that 
shall  he  to  each  other  as  3  to  4. 

Draw  through  A  a  straight  line  A  C, 
making  an  acute  angle  with  AB.  Upon 
^  C  lay  off  3  +  4  or  7  equal  parts  end- 
ing at  the  point  D.  Draw  DB.  Draw 
through  E  the  third  division  point  a  line 
EF,  parallel  to  DB,  meeting  AB  3kt  F. 

Then  AF:BF=^A. 

2.  Divide  a  straight  line  into  two 
parts  that  have  the  ratio  4:9. 

3.  Produce  a  straight  line  ^i5  to  a  point  H,  so  that 

AB:BH  =  7.5. 

4.  Divide  a  straight  line  into  three  parts  which  shall  be  to  each 
other  as  the  numbers  2,  3,  5. 

5.  Draw  a  straight  line  AB  and  then  divide  it  into  two  parts  that 
shall  have  to  each  other  the  same  ratio  as  the  two  lines  a  and  b  given 
below. 


b— 


Hint.     Draw  a  straight  line  making  an  acute  angle  with  ABy  and  lay 
off  upon  it  the  lengths  a  and  b. 


116 


FIRST   STEPS   IN   GEOMETRY 


6.  Draw  a  straight  line,  and  then  divide  it  into  two  parts  such  that 
one  part  shall  be  two  and  one  half  times  as  long  as  the  other. 

7.  Construct  a  straight  line  the  length  of  which  shall  be  the  fourth 
proportional  to  the  lengths  m,  n,  p  of  the  three  unequal  straight  lines 
(Fig.  183). 

First  Method.  Measure  the  lengths  m,  n,  p  as  closely  as  possible ; 
suppose  them  to  be  3.2  cm,  4.1  cm,  and  4.8  cm. 


Then 
Hence, 
and 


3.2:4.1  =4.8:  a;. 
3.2  x=:  4.1  x4.8, 
X  =  6.15  cm. 


Then  draw  a  straight  line  6.15  cm  long. 

This  may  be  called  the  arithmetical  solution  of  the  problem. 

Second  Method.     Draw  any  acute  angle  xAy  (Fig.  183). 

„  r*  From  A  on  Ax  take  AB  equal  to  w, 

and  BC  equal  to  n.  From  A  on  Ay 
take  AD  equal  to  p.  Draw  DB.  Draw 
CF  parallel  to  DB.  The  length  DF  is 
the  fourth  proportional  to  the  lengths 
m,  n,  p;  ot  m:n  =  p  :  DF. 

This  is  the  geometrical  solution  of 
the  problem. 

Find  by  both  the  above  methods  the 
fourth  proportional  to  the  following  lines.  Which  method  do  you 
prefer?     Give  your  reason. 


Fig.  183. 


d 
9.    ^e 


g 

10.   ■{  h 
k 


SIMILAR   FIGURES 


117 


99.  Similar  Polygons.  Similar  polygons  are  polygons  that 
have  the  same  shape. 

Let  us  first  consider  similar  triangles.  The  best  way  to 
see  clearly  their  two  most  important  properties  is  illustrated 
by  Fig.  184.  Draw  any  triangle 
ABC.  Divide  the  side  AB  into 
equal  parts ;  say  five  parts.  Through 
the  points  of  division  D,  £',  F^  G, 
draw  parallels  to  the  side  J5(7,  and 
also  parallels  to  the  side  AC. 

Now  compare  one  of  the  triangles 
thus  formed,  AEK,  with  the  triangle 
ABC.  It  has  the  same  shape  as  ABC, 
and  is,  therefore,  similar  to  ABC; 
and  it  would  remain  similar  to  ABC 
if  it  were  removed  to  any  other 
position. 

Compare  the  angles  of  the  two  triangles.  The  angle  A  is 
common.  The  angles  AEK  and  ABC  are  equal  because  they 
are  exterior-interior  angles  of  parallel  lines  (p.  54).  The 
angles  AKJE  and  ACB  are  equal  for  the  same  reason.  Thus, 
the  angles  of  the  two  triangles,  taken  pair  hy  pair  in  the  same 
order,  are  equal;  that  is,  the  two  triangles  are  equiangular 
with  respect  to  each  other. 

Examine  next  the  sides,  and  call  the  sides  opposite  the 
equal  angles  homologous  sides.  There  are  three  pairs  of 
homologous  sides:    AE  and  AB,  AK  and  AC,  EK  and  BC. 

Now  AE'.AB=2'.^',  AK:AC=2:b', 

and  EK:BC  =  2:B. 

Therefore,  AE:  AB  =  AK:  AC  =  EK:  BC. 

That  is,  the  homologous  sides  are  proportional. 


118         .  FIRST   STEPS   IN   GEOMETRY 

Thus,  it  appears  that  two  similar  triangles  have  the 
following  properties  : 

1.  They  are  equiangular  with  respect  to  each  other. 

2.  Their  homologous  sides  are  proportional. 

It  can  be  proved  that  if  two  triangles  have  either  one  of 
these  properties  they  have  the  other  also  and  are  similar. 

If,  for  instance,  we  have  two  triangles  which  are  equi- 
angiilar  with  respect  to  each  other,  they  are  similar ;  and  if 
we  have  two  triangles  which  have  their  homologous  sides 
proportional,  they  are  similar. 

EXERCISES 

Compare  the  following  pairs  of  triangles  in  Fig.  184,  showing  that 
they  have  properties  1  and  2  above  stated,  and  find  the  common  ratio 
of  their  homologous  sides  : 

1.  Triangles  ADH  and  ABC. 

2.  Triangles  ADH  and  ^Gilf. 

3.  Triangles  AEK  and  AFL. 

4.  The  sides  of  a  triangle  are  4  cm,  7  cm,  and  9  cm  in  length.  In  a 
similar  triangle  the  side  homologous  to  the  side  4  cm  long  is  12  cm 
long.     Find  the  length  of  the  other  two  sides. 

5.  The  legs  of  a  right  triangle  are  6  cm  and  8  cm  long.  A  similar 
right  triangle  has  its  shorter  leg  30  cm  long.  What  is  the  length  of  its 
other  leg  and  of  its  hypotenuse  ?  Find  the  ratio  of  the  perimeters  of 
the  two  triangles. 

6.  Draw  any  triangle  and  then  construct  a  similar  triangle,  each  side 
of  which  shall  be  one  fourth  as  long  as  the  homologous  side  of  the  first 
triangle.     What  is  the  ratio  of  the  perimeters  ? 

7.  Draw  any  triangle  and  then  construct  a  similar  triangle  each  side 
of  which  shall  be  three  fourths  as  long  as  the  homologous  side  of  the 
first  triangle.     What  is  the  ratio  of  the  perimeters  ? 

8.  Draw  a  triangle  ABC,  and  a  straight  line  DE.  Then,  taking 
DE  as  homologous  to  AB,  construct  a  triangle  DEF  similar  to  the 
triangle  ABC. 


SIMILAR   FIGURES  119 

9.  Construct  a  straight  line,  the  length  of  which  shall  be  the  mean 
proportional  between  the  lengths  of  two  given  straight  lines,  a  and  b 
(Fig.  185) ;  in  other  words,  find  a  length  c,  such  that 

a  :  c  =  c  '.h,  or  c^  =  ah. 

Solution.  Draw  a  straight  line  and  lay  ofE  upon  it  AB  equal  to  a,  and 
BC  equal  to  b.     Upon  ^  C*  as  a  diameter 

describe   a   semicircle.      At  the   point  B  _a 

erect  a  perpendicular  meeting  the  semi-  _b 

circumference  at  i).  -       D 

BD  is  the  length  required.  ^x'"''         ^   ^ 

Proof.     The  triangles  ABD,  CBD  have  /            ^^ 

right  angles  at  B^  and  the  angles  BAD  and  /         y' 

BBC  are  equal,  because  each  is  equal  to  ;'  y^ 

90°  -  ABB.     Therefore,  the  third  angles  a "o — B 

BBA   and  BCB  are   equal.     Hence,  the  Fig.  185. 
triangles  are  similar. 

Therefore,  AB  .  BB  =  BB  :  BC,  or  B&  =  AB  x  BC. 
That  is,  if  BB  is  denoted  by  c,  c^  =  a  x  b. 

In  general,  the  mean  proportional  between  two  numbers  is  a  number 
whose  square  is  equal  to  their  product.  For  example,  the  mean  propor- 
tional between  4  and  9  is  6. 

10.  Find  the  mean  proportional  between  9  and  16. 

11.  Find  the  mean  proportional  between  16  and  25. 

12.  Draw  a  rectangle,  and  then  construct  a  square  equivalent  to  the 
rectangle. 

Hint.  Find  the  mean  proportional  between  the  adjacent  sides  of  the 
rectangle. 

13.  Draw  a  parallelogram,  and  then  construct  a  square  equivalent  to  it. 
Hint.   Find  the  mean  proportional  between  the  base  and  altitude  of 

the  parallelogram. 

14.  Draw  a  triangle,  and  then  construct  a  square  equivalent  to  it. 
Hint.     Find  the  mean  proportional  between  the  altitude  and  half 

the  base. 

15.  Draw  a  trapezoid,  and  then  construct  a  square  equivalent  to  it. 
Hint.    Find  the  mean  proportional  between  the  altitude  and  half  the 

sum  of  the  bases. 


120 


FIRST   STEPS   IN   GEOMETRY 


100.    Any  two  similar  polygons  (Figs.  186,  187)  satisfy 
the  two  following  conditions: 

1.  For  every  angle  in  one  of  the  polygons  there  is  an  equal 
angle  in  the  other. 

2.  The  homologous  sides  of  the  polygons  are  proportional. 
The  equal  angles,  pair  by  pair,  are  called  corresponding  or 

homologous  angles ;  and  the  sides  which  lie  between  the 
vertices  of  homologous  angles,  pair  by  pair,  are  called  corre- 
sponding or  homologous  sides.  Thus  the  angles  UAB  and 
E'A'B',  B  and  B',  BCD  and  J5'C"i)'  are  homologous  angles, 
and  the  sides  AB  and  A'B',  BC  and  B'C,  CD  and  CD'  are 
homologous  sides. 


Fig.  186. 


Fig.  1ST. 


Two  polygons  may  be  equiangular  with  respect  to  each 
other,  and  yet  not  be  similar;  a  square  and  a  rectangle,  for 
example,  are  not  similar.  Likewise,  two  polygons  may  have 
their  sides,  taken  in  the  same  order,  proportional,  and  yet  not 
be  similar;  a  square  and  a  rhombus,  for  example,  are  not 
similar.  Before  we  can  say  that  two  polygons  are  similar, 
we  must  be  sure  that  they  satisfy  each  of  the  two  conditions 
mentioned  above. 

Triangles  are  unlike  other  polygons  in  this  respect.  If 
two  triangles  satisfy  either  of  the  two  conditions,  they  satisfy 
the  other  condition,  and  are  similar. 


SIMILAR   FIGURES 


121 


101.    It  can  be  proved  that 

Two  polygons   composed  of  the  same   number  of  triangles^ 
similar  pair  by  pair  and  similarly  placed^  are  similar. 


EXERCISES 

1.  Construct  a  polygon  similar  to  a  given  polygon  ABC  BE  (Fig, 
186),  having  given  the  side  A^B'  corresponding  to  AB. 

Solution.     Draw  the  diagonals  AC  and  AD. 

Construct  the  triangle  A'B'C  similar  to  the  triangle  ABC.  Then 
construct  the  triangle  A' CD'  similar  to  the  triangle  A  CD,  and  the 
triangle  A'D'E'  similar  to  the  triangle  ADE.  The  polygon  A'B'C'D'E' 
is  the  polygon  required. 

2.  Construct  a  quadrilateral  similar  to  a  given  quadrilateral  A  BCD 
(Fig.  188),  having  each  of  its  sides  three  fourths  as  long  as  the  correspond- 
ing side  of  the  given  quadrilateral. 

Draw  the  diagonal  AC. 

On  AB  lay  oft"  a  length  AE  equal 
to  three  fourths  of  AB. 

Draw  EF  parallel  to  BC,  meeting 
AC  Sit  F.  Draw  FG  parallel  to  CD, 
meeting  AD  at  G.  The  quadrilateral 
AEFG  is  the  quadrilateral  required. 

3.  Construct  a  square,  and  then 
construct  a  square  each  side  of  which 
shall  be  two  thirds  as  long  as  a  side 
of  the  first  square. 

4.  Construct  two  similar  rectangles,  one  with  sides  4  cm  and  7  cm 
long,  the  other  with  sides  twice  as  long. 

5.  Construct  two  similar  rhombuses,  having  for  angles  45°  and  135°; 
and  for  homologous  sides  6  cm  and  12  cm. 

6.  Construct  three  similar  parallelograms,  having  for  angles  60°  and 
120°  ;  and  their  homologous  sides  proportional  to  the  numbers  1,  2, 
and  3. 

7.  Construct  two  similar  isosceles  trapezoids  so  that  their  correspond- 
ing sides  shall  have  the  ratio  1  to  2. 

What  is  the  ratio  of  their  perimeters  ? 


Fig.  188. 


122 


FIRST    STEPS   IN    GEOMETRY 


Fig.  189. 


102.  Areas  of  Similar  Figures.  If  the  side  of  one  square 
is  double  the  side  of  another  square,  the  ratio  of  their  areas 
is  1:4;   that  is,  as  the  square  of  1  is  to  the  square  of  2. 

If  the  ratio  of  the  sides  of  two 
squares  is  2:3,  the  ratio  of  their 
areas  is  4 :  9  (Fig.  189). 

That  is,  the  areas  of  two  squares 
are  to  each  other  in  the  same  ratio 
as  the  squares  of  the  lengths  of 
their  sides. 

Consider  the  similar  triangles 
ABC^  ADE  (Fig.  190).  Their  homologous  sides  are  to  each 
other  as  1:2.  But  their  areas  are  as  1:4;  this  is  evident 
if  we  divide  the  triangle  ADE 
into  four  equal  triangles  as 
shown  in  the  figure. 

Consider  the  similar  triangles 
ABC^  AFG.  Their  homologous 
sides  are  to  each  other  as  1:3; 
their  areas  are  as  1  :  9. 

Consider  the  similar  triangles 
ADE,  AFG. 

Their  homologous  sides  are 
to  each  other  as  2:3. 

Their  areas  are  to  each  other  as  4 :  9. 

Again,  the  homologous  sides  of  the  similar  triangles  ADE, 
AHK  are  to  each  other  in  the  ratio  of  2:4,  and  their  areas 
are  in  the  ratio  of  4  :  16.     It  follows,  therefore,  that 

The  areas  of  two  similar  triangles  are  to  each  other  as  the 
squares  of  any  two  homologous  sides.     In  general, 

The  areas  of  two  similar  polygons  are  to  each  other  as  the 
squares  of  two  homologous  sides. 


Fig.  190. 


SIMILAR   FIGURES  123 


EXERCISES 


1.  The  homologous  sides  of  two  similar  triangles  are  2  ft.  and  100  ft. 
The  area  of  the  smaller  triangle  is  3  sq.  ft.  What  is  the  area  of  the 
larger  triangle? 

2.  One  side  of  a  triangle  is  5  ft.  long.  Find  the  homologous  side  of 
a  similar  triangle  twenty-five  times  as  large. 

3.  The  sides  of  a  triangle  are:  AB  =  4:00  ft.,  .4C=500  ft.,  BC  = 
600  ft.  The  triangle  is  to  be  divided  into  two  equivalent  parts  by  a 
straight  line  DE  parallel  to  BC. 

Find  the  lengths  oi  AD  and  AE.  A 

Hint.  The  triangles  ADE  and^^C 
(Fig.  191)  are  similar,  and  their  areas 
are  as  1  to  2. 


Therefore,  AD' :  AB"  =  1:2. 

„           .J.      AB      ABy/2 
Hence,  AD  =  ——  = 

=  AB  X  0.707. 

4.  Explain  how  to  draw  a  line  parallel  to  one  side  of  a  triangle  so 
as  to  cut  off  a  smaller  triangle  equal  in  area  to  one  fourth  of  the  given 
triangle. 

103.  Any  two  equilateral  triangles  have  the  same  shape 
and  are  similar  figures.  The  same  is  true  of  any  two  squares. 
It  can  be  proved  that 

Ani/  two  regular  polygons  having  the  same  number  of  sides 
are  similar  figures. 

EXERCISES 

1.  The  sides  of  two  equilateral  triangles  are  2  ft.  and  12  ft.  in  length. 
What  is  the  ratio  of  their  areas  ? 

2.  What  is  the  easiest  way  to  construct  a  regular  pentagon  one  fourth 
as  large  as  a  given  regular  pentagon  ? 

3.  Two  parks  have  the  shape  of  regular  hexagons.  If  their  sides  are 
1000  ft.  and  8000  ft.,  respectively,  what  is  the  ratio  of  their  areas? 
What  is  the  ratio  of  their  perimeters? 


124 


FIRST   STEPS   IN   GEOMETRY 


104.  Drawing  to  Scale.  In  drawing  to  scale,  the  properties 
of  similar  figures  are  usefully  applied.  A  plan  of  a  field,  for 
example,  is  a  figure  exactly  resembling  the  field  in  shape,  but 
much  smaller  in  size.  It  is  a  reduced  copy  of  the  field.  It 
is  made  by  reproducing  the  angles  of  the  field  without  change, 
but  reducing  all  lines  in  a  fixed  ratio. 

Thus,  if  a  line  on  the  ground  160  m  long  is  represented  by 
a  line  0.16  m  or  16  cm  long,  a  line  on  the  ground  800  ft. 
long  must  be  represented  by  a  line  0.8  ft.  or  9.6  in.  long,  and 


so  on. 


A  common  mode  of  reducing  lengths  in  a  given  ratio  is  by 
means  of  a  plotting  scale.  A  portion  of  a  plotting  scale,  divided 
decimally,  is  shown  in  Fig.  192. 


P  9  8 

7  6  S 

4 

3  2 

1  ( 

9                                   ^1                                     2                                    3 

M 

1 

) 

2 

1     ^          1 

(1 

3 

r 

1                I 

4 

5 

IT  T      9 

6 

P 

7 

qziiitlT 

8 

m 

1     1 

9 

i  ,luil 

10 

c 

Fig.  192. 


D 


E 


This  scale  is  constructed  as  follows.  Draw  a  straight  line 
and  take  on  it  equal  parts  AB^  BC,  CD,  etc.  Let  each  part 
represent  100  units  of  length  (feet,  meters,  etc.).  Erect  per- 
pendiculars at  the  points  J,  B,  C,  etc.  Draw  ten  straight 
lines  parallel  to  AD,  and  equidistant  from  one  another,  as 
shown  in  the  figure.  Divide  AB  and  FO  each  into  ten  equal 
parts.  Join  the  points  of  division  of  OF  to  those  of  AB  by 
lines  drawn  obliquely  as  shown. 


SIMILAR   FIGURES  125 

Number  the  points  of  division  between  0  and  F  from  1  to 
10 ;  also  number  the  points  of  intersection  of  OB  with  the 
parallels  in  the  same  way. 

Then  from  the  properties  of  similar  triangles  it  follows 
that  the  distance  from  the  point  1  on  0J5,  measured  on  the 
parallel  through  1,  to  the  first  oblique  line  is  equal  to  1  unit 
of  length ;  the  distance  from  1  to  the  second  oblique  line  is 
equal  to  11  units  of  length ;  the  distance  from  2  to  the  first 
oblique  line  is  equal  to  2  units  of  length  ;  the  distance  from 
2  to  the  second  oblique  line  is  equal  to  12  units  of  length; 
and  so  on. 

The  point  0  is  the  zero  of  the  scale;  hundreds  are  read 
off  to  the  right,  tens  to  the  left,  and  units  on  the  vertical 
line  OB. 

From  this  explanation  it  is  easy  to  see  that  a  length  of 
348  m  measured  on  the  ground  is  represented  on  the  scale 
by  the  line  mn ;  that  the  length  pq  corresponds  to  a  distance 
on  the  ground  of  216  m;  the  length  rs,  to  183  m,  etc. 

EXERCISES 

1.  If  the  scale  of  reduction  is  4  in.  to  1  mile,  what  lengths  on  paper 
will  represent  an  actual  length  of  4  miles?  2^  miles?  \  mile?  1320  ft.? 
660  ft.?  880  yd.?  2000  ft.? 

2.  If  the  scale  of  reduction  of  a  map  is  known  to  be  2  in.  to  the 
mile,  and  the  distance  between  two  cities  on  the  map  is  6|^  in.,  what  is 
the  actual  distance  between  the  cities  ? 

3.  Construct  with  the  help  of  the  scale  in  Fig.  192  a  triangle  the 
lengths  of  whose  sides  are  137  m,  160  m,  and  225  m. 

4.  A  certain  field  ABODE  has  the  shape  of  a  pentagon.  Its  angles 
are:  A  90°;  B  100°;  C  120°;  D  140°;  E  90°.  The  lengths  of  three 
sides  are:  AB  417  m;  BC  500  m;  CD  600  m.  Construct  a  plan  of 
this  field  with  the  aid  of  the  scale  in  Fig.  192  and  a  protractor.  Find 
from  the  scale  the  length  of  the  sides  DE  and  EA . 


126  FIRST    STEPS   IN    GEOMETRY 

105.  Indirect  Measurement  of  Lengths.  When  we  measure 
the  length  of  a  line  on  paper  with  a  graduated  ruler,  or  a  dis- 
tance on  the  ground  with  a  chain,  we  are  said  to  use  direct 
measurement.  But  it  often  happens  that  direct  measure- 
ment is  inconvenient  or  even  impossible,  as  in  the  case  of 
the  distance  from  the  earth  to  the  moon. 

In  these  cases  methods  of  indirect  measurement  are  used. 
We  measure  directly  certain  lines  and  angles  which  are  so 
related  to  the  line  to  be  measured  that  its  length  can  be 
computed  from  our  measurements. 

Some  problems  on  indirect  measurement  will  now  be 
considered. 

The  instruments  in  common  use  for  measuring  lengths  are  the 
engineer's  chain,  100  ft.  long,  divided  into  100  links;  the  tape  measure, 
usually  50  ft.  long,  divided  into  feet  and  inches ;  the  Gunter's  chain, 
G6  ft.  long,  divided  into  100  links. 

For  measuring  angles,  surveyors  and  engineers  employ  costly  instru- 
ments called  transits  and  theodohtes.  But  in  school  work,  a  rough 
degree  of  accuracy  will  answer  the  purpose,  and  an  angle  on  the  ground 
may  be  immediately  constructed  without  the  aid 
of  a  protractor  by  the  pin-ruler  method  (Fig.  193). 
Suppose  we  wish  to  transfer  to  paper  the  angle 
formed  at  a  certain  spot  by  lines  drawn,  one  to  a 
house,  the  other  to  a  tree.  Fasten  a  sheet  of  paper 
to  a  small  drawing  board,  and  mount  the  board  in 
a  horizontal  position  directly  over  the  given  spot, 
using  for  this  purpose  a  tripod  and  a  spirit  level. 
Stick  a  pin  into  the  paper  at  C  and  hold  the  ruler 
•against  it.  Let  one  person  sight  along  the  edge  AB  of  the  ruler,  while 
another  person  holds  a  second  pin  against  the  edge  and  turns  the  ruler 
about  C  till  the  line  of  feight  strikes  the  house. 

Prick  a  hole  D  in  the  paper  with  the  second  pin.  The  line  CD  on 
the  paper  is  one  side  of  the  required  angle.  Turn  the  ruler  about  C, 
still  keeping  the  second  pin  touching  the  edge  AB,  till  the  line  of  sight 


Fio.  193. 


SIMILAR   FIGURES 


127 


passes  through  the  tree.     Prick  a  hole  E  with  the  second  pin.     Tlie 
angle  DCE  on  th.e  paper  is  the  angle  formed  by  the  house  and  the  tree. 

106.  Problem  i.  To  yneasure  a  line  the  ends  of,  ivhicK  only 
are  accessible. 

Let  AB  (Fig.  194)  be  the  line  to  be  measured.  Choose  a 
point  C  from  which  A  and  i>  are  visible.  Measure  AC  and 
BC.  Construct  on  paper  by  the  pin-ruler  method  an  angle 
ach  =  ACB  (Fig.  195).  Lay  off  to  a  suitable  scale  of  reduc- 
tion the  reduced  lengths  ac  and  he  of  the  lines  AC  and  BC. 


Fig.  194. 


Fig.  195. 


Draw  ah  and  measure  this  line  carefully.  The  length  of 
AB  corresponding  to  ah  is  the  length  required. 

If  the  angle  ACB  is  measured  with  a  transit,  an  equal 
angle  ach  must  be  constructed  on  paper  with  a  protractor. 

Note.  The  use  of  a  figure  on  paper  may  be  avoided  by  constructing 
a  triangle  CDE  on  the  ground  similar  to  CAB.  Extend  AC  to  D  and 
BC  to  E,  each  by  the  same  fraction  of  its  length. 


EXERCISE 

1.    If  AC  (Fig.  194)  =  1000  ft.,  BC  =  800  ft.,  ah 
scale  is  1  in.  to  200  ft,  find  A  B. 


Q^^  in.,  and  the 


128 


FIRST    STEPS   IN    GEOMETRY 


107.    Problem  2.      To  measure  a  line  of  which  one  end  only 

is  accessible. 

Measure  a  straight  line  AC  (Fig.  196)  to  a  point  C  from 
which  B  is  visible.  Choose  a  scale  of  reduction,  find  the 
j-educed  length  ac  of  AC,  and  draw  ac  on  paper.  Construct 
by  the  pin-ruler  method  the  angles  hac  =  BAC,  and  hca  =  BCA. 
Complete  the  triangle  abc  on  the  paper.  Measure  with  care 
ah,  and  then  find  the  length  AB  corresponding  to  ab. 


^ 

B 

^^Q^..?~    ■ 

M>/. 

-\ 

■ ^^>^>^^?='        *e 

\^      • 

ri 

N 

mi,,.. 

..i(/(. 

Fig. 196. 


Fig.  197. 


Another  Method.  The  measurement  of  any  other  angles 
than  right  angles  may  be  avoided  by  proceeding  as  follows : 
Starting  at  A  (Fig.  197)  lay  off  a  straight  line  AC  per- 
pendicular to  AB.  Draw  at  C  a  perpendicular  C£J  to  AC. 
Bisect  AC  at  I).  Find  on  the  line  C£J  the  point  E,  which  is 
in  the  straight  line  BD  extended.  Measure  CJ^.  The  length 
of  CB  is  equal  to  the  distance  from  A  to  B.  For  the  right 
triangles  ABD  and  CIW  are  equal  (p.  63). 


EXERCISES 

1.  Find  AB  (Fig.  19G)  if  A  C  =  1200  ft.,  BA  C  =  100°,  A  CB  =  50°. 

2.  Explain  why,  if  we  lay  o^  AD  (Fig.  197)  perpendicular  to  AB 
and  of  such  length  that  the  angle  ADB  =  45°,  we  know  that  AB  =  AD. 


SIMILAR   FIGURES 


129 


108.    Problem  3.      To  measure  a  line  ivliolly  inaccessible. 

Let  it  be  required  to  find  the  length  of  the  island  AB 
(Fig.  198)  by  measurements  made  on  the  mainland.  Choose  a 
level  place,  lay  off  a  straight  line  CD  and  measure  its  length. 
Lay  off  on  paper  the  reduced  length  cd.  Construct  by  the 
pin-ruler  method  the  angles  acb=ACB,  hcd=BCD^  adh=ADB^ 
adc  =  ADC.  Draw  ah.  Measure  this  line.  The  distance 
AB  is  the  length  corresponding  to  ab.  The  number  of  arglc  s 
to  be  constructed  in  applying  this  method  makes  it  difficult 
to  secure  a  good  result. 


i 

^^^^ 

D  ■*• 

^^3 

^ 

^ 

S^M 

Fig.  198. 


Fig. 199. 


Another  Method.  Let  it  be  required  to  measure  the  dis- 
tance from  ^  to  i>  (Fig.  199)  without  crossing  the  river. 

Lay  off  a  straight  line  MN.  Find  on  this  line  the  points 
C  and  Z>  through  which  perpendiculars  from  A  and  B,  respec- 
tively, will  pass.  Bisect  CD  at  ^.  Extend  AE  to  meet  the 
perpendicular  BD  at  F,  and  extend  BF  to  meet  the  perpen- 
dicular ^C  at  6^.     Then  AC  =  DF,  BD  =  CG,  and  AB  =  GF. 

Nothing  remains,  therefore,  to  be  done  but  to  measure  GF. 

In  this  method  right  angles  only  are  used,  and  the  laws 
of  equal  triangles  are  applied.  Wliat  pairs  of  triangles  in 
Fig.  199  are  equal  ?  Why  are  they  equal  (see  p.  63)  ?  What 
kind  of  a  figure  is  ABFG  ? 


130 


FIRST   STEPS   IN   GEOMETRY 


109.  Problem  4.  To  iiieasure  the  height  of  an  object  standing 
on  a  horizontal  plane. 

Let  AB  (Fig.  200)  be  the  object.  From  the  foot  A  of  the 
object  measure  in  any  direction  a  straight  line  AC.  Place 
the  instrument  for  measuring  angles  so  that  its  centre  shall 
be  at  D,  directly  above  C,  and  its  line  of  sight  shall  be  hori- 
zontal, meeting  AB  at  E.  Measure  DC  and  the  angle  EDB., 
which  is  called  the  angle  of  elevation  of  the  point  B.  Draw 
to  scale  a  right  triangle  edh  similar  to  EDB.  Measure 
eh  and  find  EB^  the  length  corresponding  to  the  reduced 
length  eh.     Then  EB  +  DC  =  AB. 

B 


<:^':f.e 


Fig.  200. 


X: 

c ,. 


Another  Method.  Find  the  point  F  where  BD  extended 
meets  ^C  extended.  Measure  ^i^,  (77^,  and  CD.  Then,  since 
the  triangles  AFB  and  CFD  are  similar,  CF :  CD  =  AF :  AB. 

CD  X  AF 
Whence,  AB  =  — — 777; • 


EXERCISES 

1.  If  EDB  =  30°,  AC  =  240  ft.,  DC  =  4  ft.,  find  AB. 

2.  If  yl  C  =  300  ft.,  CF  =  10  ft.,  DC  =  4  ft.,  find  A  B. 

3.  How  can  this  problem  be  soh'ed  by  using  the  angle  45°? 


SIMILAR  FIGURES 


131 


110.    Problem  5.      To  find  the  height  of  an  object  by  means 
of  its  shadow. 

When  a  vertical  object  casts  a  shadow  on  a  horizontal 
plane,  the  length  of  the  object,  the  length  of  the  shadow,  and 
the  ray  of  light  from  the  top  of  the  object  to  the  end  of  the 
shadow  form  a  right  triangle 
(Fig.  201).  Moreover,  since 
the  sun  is  so  far  away  that 
his  rays  are  sensibly  parallel 
lines,  the  right  triangles 
formed  at  an}^  instant  by 
the  sun  shining  on  two 
vertical  objects  are  similar. 
These  facts  are  applied  to 
find  the  height  of  an  object. 

Suppose  that  a  tree  AB 
(Fig.  201)  casts  the  shadow 
AC^  and  that  at  the  same  time  a  vertical  pole  BE  casts  the 
shadow  DF.  Measure  A  C,  I)E,  I)F.  Then,  since  the  triangles 
ABC  and  DEF  are  similar,  ABiDE  =  AC:  DF. 

ACXDE 

Whence,  AB  = =77, 

'  DF 


Fig.  201. 


EXERCISES 

1.  If  (Fig.  201)  ^C  =  60  ft.,  DE  =  10  ft.,  DF  =  S  ft.,  find  AB, 

2.  JiAC=  70  ft.,  and  the  angle  DFE  =  45°,  find  AB.y 

3.  The  shadow  of  a  church  spire  is  120  ft.  long,  and  at  the  same 
time  the  shadow  of  a  man  6  ft.  tall  is  3^  ft.  long.  Find  the  height  of 
the  spire. 

4.  How  high  is  a  monument  the  shadow  of  which  is  24  m  long, 
when  at  the  same  time  a  meter  rod  casts  a  shadow  60  cm  long  ? 

5.  If  the  shadow  of  a  vertical  rod  is  half  as  long  as  the  rod,  what  is 
the  height  of  a  tree  the  shadow  of  which  is  80  ft.  long? 


132  FIRST   STEPS   IN   GEOMETRY 

111.    Problem  6.      To  find  the  height  of  an  inaccessible  object. 

Let  it  be  required  to  find  the  height  AB  (Fig.  202)  of  a 
tree  on  the  farther  bank  of  a  river  without  crossing  the  river. 
Find  the  point  C  directly  opposite  the  tree  where  the  angle  of 
elevation  ACB  of  the  tree  is  45°.  Lay  off  from  C  a  straight 
line  CD  perpendicular  to  AC  and  of  such  length  that  the 
angle  ADC  =  45°.  Then  we  know  that  AB  =  AC  =  CD.  All 
that  remains  to  be  done  is  to  measure  the  length  of  CD. 


''■■^''t.^% :} 


^^^i-ii^^*^^^ 


V      ^    ■-.   .....  *■■■ ■-•(7'''^"-~v '•■■=::■'■■•■-: 


Fig.  202. 

Those  who  are  familiar  with  the  pin-ruler  method  of  con- 
structing angles  (p.  126)  will  easily  perceive  how  the  angles 
ACB  and  ADC  may  be  made  equal  to  45°  by  means  of  a 
square  drawing  board  with  its  diagonals  drawn. 

EXERCISE 

1.  Wishing  to  find  the  height  AB  oi  a,  tree,  I  observe  the  angle  of 
elevation  A  CB  of  the  top  B  as  seen  from  a  point  C,  and  find  it  to 
be  30°.  At  a  point  D  on  the  line  A  C  and  120  ft.  nearer  the  tree  I 
find  the  angle  of  elevation  ADB  to  be  45°.  Find  the  height  AB 
of  the  tree. 


SIMILAR  FIGURES  133 

112.  Surveying.  To  survey  a  piece  of  land  is  to  make  such 
measurements  as  will  determine  its  size  and  shape,  to  com- 
pute its  area,  and  to  draw  to  scale  a  plan  of  the  land. 

The  only  instruments  required  for  making  the  measure- 
ments on  the  ground  are  a  Gunter's  chain  for  measuring 
lengths,  and  a  square  drawing  board  with  its  diagonals  drawn 
for  constructing  right  angles  and  running  perpendiculars. 

A  Gunter's  chain  is  4  rods  or  66  ft.  long  and  contains  100 
links.  It  has  two  advantages:  first,  the  number  of  links 
can  be  written  decimally  as  hundredths  of  a  chain  ;  secondly, 
square  chains  can  be  reduced  to  acres  by  dividing  by  10. 

1  acre  =  160  sq.  rods  =  10  X  16  sq.  rods  =  10  sq.  chains. 

Problem  i.     To  survey  the  open  four-sided  field  ABCD. 

Begin  by  marking  out  and  measur- 
ing the  diagonal  BD  (Fig.  203). 

Then  determine  the  points  E^  F  on 
BD^  where  perpendiculars  from  the 
points  A  and  C,  respectively,  meet 
BD.  Then  measure  the  lengths  of 
AE  and  CF.  Then  find  the  areas 
of  the  triangles  ABD  and  CBD. 
Their  sum  will  be  the  area  of  the  p^^  203. 

field.     Lastly,  draw  to  scale  a  plan 
of  the  field,  beginning  with  the  diagonal  BD. 


EXERCISE 

1.   Find  the  area  and  construct  to  a  suitable  scale  a  plan  of  the  field 
ABCD  from  the  following  measurements : 

BD  -  14.36  chains,  AE  =  8.17  chains,  CF  =  5.74  chains. 


134 


FIRST   STEPS   m   GEOMETRY 


Problem  2.      To  survey  a  field  ABCD  when  its  diagonals 
cannot  conveniently  he  7neasured. 

Through  one  corner  A  of  the  field  run  a  straight  line 
EAGF.  Find  the  points  E,  G,  F  where  perpendiculars  from 
jr^ ^^C  the  corners  B^  C,  and  D,  respec- 
tively, meet  the  line.  Measure 
AE,  BE,  AF,  CF,  AG,  and  J)G. 
Find  the  areas  of  the  trapezoids 
EBCF  and  GDCF;  and  of  the  tri- 
angles AEB  and  AGI),  Find  the 
area  of  the  field,  and  draw  to  scale 
a  plan  of  the  field,  erasing  all  the 
auxiliary  lines. 

Fig.  204. 


EXERCISES 
1.    Find  the  area  of  the  field  in  Fig.  205,  having  given 

AF  =  12  chains,  BF  =    7  chains,  A G  =  8  chains,  CG  =  12  chains, 
AK  =    3  chains,  DK  =  12  chains,  AL  =8  chains,  EL  =    5  chains. 


Fig.  206. 


2.  Describe  how  you  would  proceed  in  order  to  find  the  area  of  a 
field  having  the  shape  ABODE  (Fig.  206).  Begin  by  drawing  a  figure 
similar  to  ABODE.  Then  put  in  the  lines  which  you  would  measure 
in  order  to  obtain  the  area.  Then  describe  clearly  how  you  would 
compute  the  area. 


SIMILAR   FIGURES  136 

Problem  3.  To  survey  an  inaceessible  field  ABCD  by 
means  of  an  auxiliary  figure. 

Sometimes  the  direct  measurement  of  straight  lines  in  a 
field  is  difficult  or  impossible   by  reason  of  obstacles  such 
as  woods,  swampy  land,  etc.     One  way 
of  meeting  this  difficulty  is  illustrated  ,'\ 

in  Fig.  207. 

First,  measure  the  lengths  of  the  sides  /        \ 

AB,  BC\  CD,  and  DA.  Then  extend 
the  sides  AD  and  BC  till  they  meet  at 
E.  Then  draw  to  scale  a  plan  dee  of 
the  triangle  DCK  Extend  on  the  paper 
ed  to  a  point  a  such  that  the  length  ad 
on  the  paper  shall  correctly  represent, 
to  the  scale  used,  the  length  AD  on  the 

gi'ound.  Extend  in  the  same  way  ee  to  a  point  b  so  that 
bo  on  the  paper  shall  represent  ^C  on  the  ground.  Draw  ab. 
The  figure  abed  on  the  paper  is  a  plan  of  the  field  ABCD. 

To  find  the  area  of  the  field,  first  find  the  area  of  the  figure 
abed,  then  multiply  this  area  by  the  square  of  the  number  of 
chains  represented  by  1  unit  of  length  on  the  plan ;  the 
product  will  be  the  area  of  the  field  in  square  chains. 

For  example,  suppose  that  the  scale  of  reduction  is  1  cm 
to  3  chains.  Then  1  qcm  of  area  on  the  plan  represents  an 
actual  area  on  the  field  of  9  sq.  chains.  If  the  area  of  abed 
is  12.27  qcm,  then  the  area  of  the  field  will  be  12.27  X  9  = 
110.43  sq.  chains  =  11.043  acres. 

EXERCISE 

1.  If  (Fig.  207)  AB  =  24:  chains,  BC  =  12  chains,  DC  =  10.58  chains, 
AD  =  IQ  chains,  DE  =  8  chains,  CE  =  12  chains,  construct  a  plan  of 
the  field  and  find  its  area. 


136 


FIRST   STEPS   IN   GEOMETRY 


Another  Metliod,  Lay  out  a  rectangle  which  completely 
encloses  the  field,  and  draw  perpendiculars  from  all  the 
corners  of  the  field  to  the  sides  of  the  rectangle.  Subtract 
the  sum  of  the  areas  of  the  right  triangles  and  trapezoids 
thus  formed  from  the  area  of  the  rectangle,  the  remainder 
will  be  equal  to  the  area  of  the  field. 


Fig.  208. 


EXERCISE 

Find  the  area  of  the  polygon  ABCDEFGIIK,  having  given 


SA  =  30  chains 
AL  =25 
LB  =20 
BC  =45 
CM  =  20 
MN  =  15 


MD  =  10  chains 
NE  =  35       " 
EO  =20 
OP  =25       " 
PQ  =  15       " 


QII  =  20  chains 
HR  =  22       " 
SR  =  18 
PF  =  10       « 
QG  =  15       " 
RK  =  12       " 


Arrange  the  results  of  computing  the  areas  in  neat  form  as  follows 

Rectangle  SLNO       =  55  x  100  =  5500  sq.  chains  =  550  acres. 
Right  triangle  .4  Zi?  =  25  X  10    =250     "        "      =25       " 


K 


;a 


^' 


CHAPTER   VII 


THE   COMMON   GEOMETRIC   SOLIDS 

113.  The  Right  Prism.  A  solid  bounded  by  two  equal 
polygons  in  parallel  planes  and  by  three  or  more  rectangles 
is  called  a  right  prism  (Fig.  209). 

The  two  polygons  are  called  the  bases,  the  distance  between 
the  bases  is  called  the  height  or  altitude,  the  rectangles  are 
called  the  lateral  faces,  the  intersections  of  the  lateral  faces 
are  called  the  lateral  edges. 

A  right  prism  is  said  to  be  triangular,  quadrangular,  etc., 
according  as  its  bases  are  triangles,  quadrilaterals,  etc. 


Fio.  209. 


Fig.  210. 


A  right  prism,  the  bases  of  which  are  regular  polygons, 
is  called  a  regular  prism. 

A  right  prism,  the  bases  of  which  are  rectangles  (Fig.  209), 
is  called  a  rectangular  parallelepiped. 

A  rectangular  parallelopiped  may  also  be  defined  as  a  solid 
bounded  by  six  rectangles. 

A  cube  is  a  solid  bounded  by  six  equal  squares  (Fig.  210). 

137 


Fig.  211. 


138  FIRST   STEPS   IN   GEOMETRY 

114.    Parallel  Perspective.     To  make  a  drawing  of  a  cube 
like  that  in  Fig.  211,  begin  by  drawing  the  square  ABCD. 
jj  Then  draw  AE  making  the  angle 

BAU  equal  to  30°  and  AE  equal 
to  1  AB.  Then  draw  BF,  CG,  and 
DH  parallel  and  equal  to  AE, 
Draw  EF,  IIG,  EH,  and  EG. 
The  edges  of  the  cube  are  now  all 
represented  on  the  paper,  and  the 
diagram  is  complete.  The  edges 
shown  with  dashes,  AE,  EF^  EH,  are 
the  edges  which  would  be  invisible  if  the  cube  were  made  of 
an  opaque  material  such  as  wood. 

In  this  diagram  the  face  ABCD  is  to  be  regarded  as  lying 
in  the  plane  of  the  paper,  the  face  EFGH  as  parallel,  and  the 
other  faces  as  perpendicular  to  the  plane  of  the  paper. 

The  eight  edges  which  lie  either  in  the  plane  of  the  paper 
or  parallel  to  it  are  drawn  in  their  true  relative  directions  and 
equal  in  length.  The  remaining  edges,  AE,  BE,  CG,  DH, 
must  be  so  drawn  that  they  shall  appear  to  be  perpendicular 
to  the  plane  of  the  paper  and  equal  to  the  other  edges ; 
the  method  of  drawing  them  just  described  is  one  way  of 
meeting  these  conditions. 

The  result  is  that  only  two  faces  of  the  cube  are  drawn 
as  squares,  the  other  faces  being  parallelograms. 

A  diagram  of  a  cube  made  as  here  described  is  called  a 
parallel  projection  or  parallel  perspective  of  a  cube. 

It  is  not  necessary  that  the  angle  BAE  in  Fig.  211  should 
be  made  equal  to  exactly  30°,  or  that  AE  should  be  made 
equal  to  exactly  \AB.  Other  values  for  the  angle  of  reduc- 
tion and  the  reducing  ratio,  within  certain  limits,  are  allow- 
able.    But  30°  and  l  are  good  values  for  the  cube. 


THE   COMMON   GEOMETRIC    SOLIDS  139 

Parallel  perspective  is  one  of  the  ways  of  representing  on 
paper  objects  that  have  three  dimensions  in  space.  It  does 
not  reproduce  the  exact  appearance  of  an  object  as  seen  from 
any  point  of  view.  It  serves,  however,  to  suggest  to  the 
mind  the  true  relations  of  the  parts  of  the  common  geometric 
solids  better  than  any  other  method.  All  lines  in  space,  for 
example,  that  are  parallel  and  equal  are  represented  in  the 
diagram  by  lines  that  are  parallel  and  equal. 

EXERCISES 

1.  Make  a  drawing  in  parallel  perspective  of  a  right  prism  witli  a 
square  base  and  lateral  edges  twice  as  long  as  one  side  of  the  base. 

This  differs  from  the  cube  only  in  the  length  of  the  lateral  edges. 

2.  Put  in  parallel  perspective  a  right  parallelopiped,  taking  as  the 
dimensions :  length  12  cm,  width  6  cm,  height  9  cm.  In  Fig.  212 
the  front  edge  of  the  base  represents  tlie 
length  of  the  parallelopiped,  and  the  width 
runs  from  the  front  backward.  In  drawing 
the  edges  that  are  perpendicular  to  the  plane 
of  the  paper  use  the  angle  of  reduction  30°, 
and  the  reducing  ratio  |,  as  in  the  case  of  ^^^  2iq 
the  cube. 

3.  Put  in  parallel  perspective  the  rectangular  parallelopiped  of  Ex.  2 
so  that  the  length  shall  extend  from  the  front  backwards. 

4.  Put  in  parallel  perspective  the  rectangular  parallelopiped  of  Ex.  2 
with  the  height  and  width  interchanged. 

5.  Draw  a  rectangular  parallelopiped,  taking  as  the  dimensions  : 
length  5  in.,  width  3  in.,  height  4  in. 

6.  Draw  a  rectangular  parallelopiped,  taking  as  the  dimensions  : 
length  8  cm,  width  4  cm,  height  10  cm. 

7.  The  dimensions  of  a  right  prism  are  :  length  8  cm,  width  G  cm, 
height  10  cm.     Put  the  prism  in  parallel  perspective, 

(1)  using  the  angle  of  reduction  30°  and  the  reducing  ratio  ^  ; 

(2)  using  the  angle  of  reduction  45°  and  the  reducing  ratio  ^  ; 

(3)  using  the  angle  of  reduction  60°  and  the  reducing  ratio  ^ . 


140 


FIRST   STEPS   IN   GEOMETRY 


8.    Put  in  parallel  perspective  a  regular  triangular  prism,  making 
its  height  one  and  a  half  times  the  length  of  one  side  of  its  base. 

Begin  by  drawing  an  equilat- 


eral triangle  and  its  altitude  (Fig. 
213).  Then  draw  a  horizontal 
line  AD  equal  to  the  altitude  of 
the  triangle.  Through  D  draw 
BC  so  that  ADB  =  30°,  EC  = 
^  AD,  and  D  is  the  middle  point 
of  BC.  Draw  AB  and  AC. 
ABC  is  the  base  of  the  prism. 
Then  draw  AE  perpendicular  to 
AD,  and  equal  to  |  of  one  side 
Draw  BF  and  CG  equal  and  parallel  to 


Fig.  213, 


of  the  equilateral  triangle. 
AE.     Draw  EF,  EG,  GE. 

9.    Draw  a  regular  hexagonal  prism,  making  its  height  three  times 
as  long  as  one  side  of  the  base. 

Draw  a  regular  hexagon  ABC  DEE  and  the  diagonals  EC,  AE,  BD. 
Then  draw  a  horizontal  line  MN  equal  to  EC.  On  this  line  lay  off 
MO  equal  to  EG,  and  NP 


equal  to  CH.  Through  0 
draw  QR  equal  to  ^  AE, 
making  the  angle  30°  with 
MN,  and  bisected  at  the 
point  0.  Through  P  draw 
aS'T' parallel  and  equal  to  QR, 
and  bisected  at  the  point  P. 
The  figure  MQSNTR  is  the 
base  of  the  prism.  Then 
draw  all  the  lateral  edges  of 
the  hexagon  perpendicular 
to  MN,  and  make  each  one 
three  times  as  long  as  QS. 


Fio.  216. 


Connect  by  straight  lines  the  upper  ends  of 
the  lateral  edges,  and  the  figure  is  complete. 

10.  Put  in  parallel  perspective  the  regular  hexagonal  prism  of  Ex.  9, 
using  the  angle  of  reduction  45°  and  the  reducing  ratio  |. 

11.  Put  in  parallel  perspective  a  right  prism  having  for  base  an  isos- 
celes triangle  with  sides  4  cm,  8  cm,  8  cm,  and  having  an  altitude  of  12  cm. 


THE   COMMON   GEOMETRIC    SOLIDS 


141 


115.  The  Regular  Pyramid.  A  solid  bounded  by  a  regu- 
lar polygon  and  three  or  more  isosceles  triangles  having  a 
common  point  for  vertex  is  called  a  regular  pyramid  (Fig.  217). 

The  regular  polygon  is  called  the  base  of  the  pyramid. 

The  triangles  are  called  the  lateral  faces  of  the  pyramid, 
and  their  intersections  are  called  the  lateral  edges. 

The  common  vertex  of  the  triangles  is  called  the  vertex  of 
the  pyramid. 

The  lateral  edges  are  equal  lines,  and  the  lateral  faces  are 
equal  triangles. 

A  pyramid  is  called  triangular,  quadrangular,  etc.,  according 
as  its  base  is  a  triangle,  a  quadrilateral,  etc. 


Fig.  217. 


Fig.  218. 


The  height  or  altitude  of  a  pyramid  is  the  length  of  the 
perpendicular  drawn  from  the  vertex  to  the  base. 

In  a  regular  pyramid  this  perpendicular  passes  through 
the  centre  of  the  base,  and  is  also  called  the  axis  of  the 
pyramid  {SP,  Fig.  218). 

The  slant  height  of  a  regular  pyramid  is  the  altitude  of  any 
one  of  the  lateral  faces.  The  slant  height  (SA^  Fig.  218) 
bisects  the  side  of  the  base  to  which  it  is  drawn.  The  slant 
height  is  always  greater  than  the  height  of  the  pyramid. 


142 


FIRST   STEPS   IN   GEOMETRY 


EXERCISES 


FiCx.  219. 


1.  Put  in  parallel  perspective  a  regular  pyramid  with  a  square  base. 
The  square  base  ABCD  (Fig.  219)  is  put  in  parallel  perspective  by 

proceeding  as  already  explained  in  the 
case  of  the  cube,  using  the  reducing  ratio 
^  instead  of  ^. 

The  axis  of  the  pyramid  is  then  drawn 
perpendicular  to  AB  from  the  point  O, 
the  intersection  of  the  diagonals  A  C  and 
BD.  From  F,  any  point  in  the  axis,  as 
the  vertex  of  the  pyramid,  draw  straight 
lines  to  the  points  A,  B,  C,  and  D. 

2.  Put  in  parallel  perspective  a  regular 
pyramid  with  a  square  base,  having  dimen- 
sions as  follows  :  side  of  base  6  cm,  height 
of  pyramid  10  cm. 

3.  Put  in  parallel  perspective  a  regular 
triangular  pyramid,  the  height  of  which  is 
twice  one  side  of  the  base. 

Construct  an  equilateral  triangle  (Fig.  220).  Draw  two  of  its  alti- 
tudes. The  intersection  of  these 
altitudes  is  the  centre  of  the  tri- 
angle, and  divides  each  altitude  into 
two  parts  such  that  the  smaller  part 
is  ^  of  the  whole  altitude. 

Draw  the  equilateral  triangle  in 
parallel  perspective  as  explained  in 
Ex.  8,  p.  140.  The  resulting  figure 
ABC  is  the  base  of  the  pyramid. 
The  foot  0  of  the  axis  of  the  pyra- 
mid is. found  by  laying  off  from  D, 
on  the  altitude,  DO  =  ^AD. 

4.  Put  in  parallel  perspective  a 
regular  hexagonal  pyramid. 

The  hexagonal  base  of  the  pyramid  is  drawn  exactly  as  explained  in 
the  case  of  an  hexagonal  prism  (Ex.  9,  p.  140). 


Fig.  220. 


Fig.  221. 


THE   COMMON   GEOMETRIC    SOLIDS 


143 


116.  Cylinder  of  Revolution.  The  solid  generated  by  a 
rectangle  revolving  about  one  of  its  sides  as  an  axis  is  called 
a  cylinder  of  revolution  (Fig.  222). 

The  side  AB  about  which  the  rectangle  revolves  is  the 
axis  of  the  cylinder ;  its 
length    is  the   height   of 
the  cylinder. 

The  opposite  side  CD 
generates  a  curved  sur- 
face called  the  lateral  sur- 
face of  the  cylinder. 

The  sides  AD  and  BC 
generate  circles  called 
the  bases  of  the  cylinder. 

A  cylinder  of  revolution  is  also  called  a  right  circular 
cylinder. 

EXERCISE 


Fig.  222. 


Pig.  223. 


H 


M 


Fig.  224. 


N 


1.  Draw  a  cylinder  of  revolution. 
If  the  axis  of  the  cylinder  is  vertical, 
the  bases  will  appear  as  ellipses. 

An  ellipse  is  drawn  with  the  redu- 
cing ratio  ^  and  without  change  of 
angles,  as  shown  in  Fig.  224.  The 
square  A  BCD  becomes  the  rectangle 
EFGH,  and  the  inscribed  circle  be- 
comes an  ellipse  inscribed  in  the  rect- 
angle and  having  for  diameters  MN 
and  PQ.  Points  on  the  ellipse  are 
found  by  drawing  chords  perpendicu- 
lar to  MN  and  reducing  each  chord. 
In  drawing  the  base  of  a  cylinder  the  square  and  rectangle  may  be 

omitted.     After  the  points  on  the  ellipse  have  been  found,  the  ellipse  is 

drawn  free-hand  through  these  points. 


A 

r 

Q 

^ 

h 

'A 

-^ 

^^ 

\ 

N 

^_ 

0 

^^ 

/ 

\ 

c 

p 

J 

V 

144 


FIRST   STEPS   IN    GEOMETRY 


117.  Cone  of  Revolution.  The  solid  generated  by  a  right 
triangle  revolving  about  one  of  its  legs  as  an  axis  is  called 
a  cone  of  revolution  (Fig.  225). 

The  leg  AB  about  which  the  triangle  revolves  is  the  axis 
of  the  cone,  its  length  the  height  or  altitude  of  the  cone. 

The  hypotenuse  BC  generates  a  curved  surface  called  the 
lateral  surface  of  the  cone. 


Fig.  225. 


Fl(i.    L'L'O. 


The  leg  A  C  generates  a  circle  called  the  base  of  the  cone. 
The  length  of  the  hypotenuse  BC  is  called  the  slant  height 
of  the  cone. 

A  cone  of  revolution  is  also  called  a  right  circular  cone. 


EXERCISES 

1.  Draw  a  cone  of  revolution. 

Begin  by  putting  a  circle  iu  parallel  perspective,  and  the  resulting 
ellipse  is  the  base  of  the  cone.  Erect  at  the  centre  of  the  ellipse  a  line, 
for  the  axis  of  the  cone. 

2.  The  height  of  a  cone  of  revolution  is  12  cm,  the  radius  of  its 
base  is  5  cm.     Find  its  slant  height  (see  p.  104). 

3.  The  slant  height  of  a  cone  of  revolution  is  20  ft.,  the  radius  of  its 
base  is  12  ft.     Find  the  height  of  the  cone. 


THE   COMMON   GEOMETRIC    SOLIDS  145 

118.  Frustums.  The  frustum  of  a  pyramid  or  of  a  cone  is 
the  portion  of  the  pyramid  or  the  cone  included  between 
the  base  and  a  plane  parallel  to  the  base  (Figs.  227  and  228). 

The  base  and  the  section  parallel  to  the  base  are  called 
the  bases  of  the  frustum.  The  height  or  altitude  of  the  frustum 
is  the  distance  between  its  bases. 


In  the  frustum  of  a  regular  pyramid  the  two  sides  of  the 
bases  contained  between  the  same  two  parallel  edges  are  par- 
allel lines;  the  lateral  faces  are  equal  isosceles  trapezoids; 
the  slant  height  is   the  altitude   of   one   of  these   trapezoids. 

In  the  frustum  of  a  cone  of  revolution  that  portion  of  the 
slant  height  of  the  cone  included  between  the  bases  of  the 
frustum  is  called  the  slant  height  of  the  frustum. 

EXERCISES 

1.  Draw  the  frustum  of  a  regular  pyramid  with  a  square  base. 
Draw  the  entire  pyramid.     Put  in  the  upper  base  of  ,f^^ 

the  frustum  by  drawing  each  one  of  its  sides  parallel  to 
the  corresponding  side  of  the  lower  base.  Erase  the 
part  above  the  upper  base. 

2.  Draw  the  frustum  of  a  regular  hexagonal  p^Tamid. 
Hint.    Begin  by  drawing  the  complete  pyramid. 

3.  Draw  the  frustum  of  a  cone  of  revolution. 
Hint.   Begin  by  drawing  the  complete  cone.  Fig.  229. 


ir\ 


146 


FIRST   STEPS   IN   GEOMETRY 


119.  The  Sphere.  A  semicircle  revolving  about  its  diam- 
eter generates  a  solid  called  a  sphere  (Fig.  231). 

The  meaning  of  the  terms  centre,  radius,  diameter,  as  applied 
to  a  sphere,  will  be  plain  from  their  definitions  as  applied  to 
a  circle. 

Every  section  of  a  sphere  made  by  a  plane  is  a  circle. 

The  circle  is  called  a  great  circle  if  the  plane  passes  through 
the  centre,  a  small  circle  in  all  other  cases  (Fig.  232). 


Fig.  230. 


Fig.  231. 


Fig.  232. 


The  part  of  a  sphere  contained  between  two  parallel 
planes  is  called  a  spherical  segment  (Fig.  232),  and  the  part 
of  the  surface  of  the  sphere  contained  between  the  planes  is 
called  a  zone.  The  circular  sections  made  by  the  planes  are 
called  the  bases  of  the  segment,  and  the  distance  between 
them  is  called  the  height  of  the  segment  or  zone. 

A  single  plane  divides  a  sphere  into  two  parts  called 
segments  of  one  base,  and  the  surface  into  two  parts  called 
zones  of  one  base.  A  great  circle  divides  a  sphere  into  two 
equal  segments  called  hemispheres. 

The  portion  of  a  sphere  generated  by  the  revolution  of  a 
circular  sector  AOB  (Fig.  232)  about  one  of  its  radii  is  called 
a  spherical  sector. 


THE    COMMON    GEOMETRIC    SOLIDS  147 

120.  Surfaces  of  the  Common  Solids.  The  most  impor- 
tant cases  are  covered  by  the  following  formulas,  in  which  S 
denotes  lateral  area  or  the  area  of  curved  surface : 

Right  Prism. 

S = perimeter  of  base  X  altitude. 

Cylinder  of  Revolution. 

aS'  =  circumference  of  base  X  altitude. 

Regular  Pyramid. 

S  =  perimeter  of  base  x  ^  slant  height. 

Cone  of  Revolution. 

S  =  circumference  of  base  X  i  slant  height. 

Frustum  of  Regular  Pyramid. 

S  =  i  sum  of  perimeters  of  bases  X  slant  height. 

Frustum  of  Cone  of  Revolution. 

S  =  ^  sum  of  circumferences  of  bases  X  slant  height. 

Sphere. 

Let  r  =  radius  of  sphere,  h  =  altitude  of  zone,  and  tt 
denote  the  number  \2^  or  3.1416. 

Area  of  surface  of  sphere  =  4  irr^. 

Area  of  zone  —  2  irrh. 


148  FIRST   STEPS   m   GEOMETRY 


EXERCISES 

1.  What  is  the  entire  surface  of  a  cube  whose  edge  is  6  cm  long  ? 

2.  How  many  sqviare  feet  of  lead  are  required  to  line  the  bottom 
and  sides  of  a  cubical  tank  4|-  ft.  deep  ? 

3.  Find  the  entire  surface  of  a  rectangular  parallelopiped  if  the 
length  =  15  cm,  the  breadth  =  9  cm,  and  the  depth  =  6  cm. 

4.  A  marble  pillar  12  ft.  high  has  a  square  base  one  edge  of  which 
is  2  ft.  4  in.  long.  What  will  it  cost  to  polish  the  lateral  surface  of  the 
pillar  at  50  cents  per  square  foot  ? 

5.  How  many  bricks  8  in.  by  3^  in.  are  required  to  line  the  bottom 
and  sides  of  a  reservoir  60  ft.  long,  24  ft.  wide,  and  12  ft.  deep? 

6.  The  height  of  a  regular  hexagonal  prism  is  32  ft.  One  side  of 
its  base  is  3  ft.     Find  its  entire  surface  (see  p.  102). 

7.  Find  the  entire  surface  of  a  right  circular  cylinder  10  in.  high, 
the  radius  of  the  base  being  5  in. 

8.  How  many  square  feet  of  tin  are  required  to  cover  the  lateral  sur- 
face of  a  circular  tower  40  ft.  high,  the  diameter  of  the  base  being  8  ft.? 

9.  Find  the  cost  of  cementing  the  bottom  and  side  of  a  cylindrical 
tank  20  ft.  deep  and  18  ft.  in  diameter  at  30  cents  per  square  foot. 

Find  the  lateral  surface  of  a  regular  hexagonal  pyramid,  given : 

10.  Edge  of  base  20  cm,  slant  height  30  cm. 

11.  Edge  of  base  10  cm,  lateral  edge  13  cm  (see  p.  104). 

12.  Edge  of  base  8  cm,  height  of  pyramid  25  cm. 

13.  How  many  square  feet  of  tin  are  required  to  cover  the  lateral 
surface  of  a  square  pyramid  16  ft.  high  and  side  of  base  8  ft.? 

14.  The  height  of  a  right  cone  is  16  cm  and  the  radius  of  its  base 
is  6  cm.     Find  its  slant  height  and  lateral  surface. 

15.  A  right  triangle  whose  legs  are  3  in.  and  4  in.  generates  a  cone 
by  revolving  about  its  shorter  leg.     Find  the  entire  surface  of  this  cone. 

16.  How  much  canvas  is  required  to  make  a  conical  tent  80  ft.  high 
and  70  ft.  in  diameter  at  the  base  ? 

17.  Find  the  total  surface  of  a  frustum  of  a  square  pyramid  if  the 
sides  of  its  bases  are  12  in.  and  4  in.,  and  the  slant  height  is  5  in. 

18.  Find  the  height  of  the  frustum  in  Ex.  17. 

19.  A  frustum  of  a  regular  triangular  pyramid  is  1 0  ft.  high,  and  the 
sides  of  its  bases  are  40  ft.  and  20  ft.     Find  its  lateral  surface. 


EXERCISES   ON   SURFACES   OF   SOLIDS  149 

20.  A  church  spire  has  the  shape  of  a  frustum  of  a  regular  hexagonal 
pyramid.  Each  side  of  the  lower  base  is  10  ft.  long,  and  each  side  of  the 
upper  base  is  4  ft.  long.  The  slant  height  of  the  frustum  is  80  ft.  How 
many  square  feet  of  tin  are  required  to  cover  the  lateral  faces  and  the  top  ? 

21.  Find  the  entire  surface  of  the  frustum  of  a  right  cone,  given  the 
radii  of  its  bases  14  cm  and  9  cm,  and  its  height  12  cm. 

22.  Find  the  surface  of  a  sphere  whose  diameter  is  7  in. 

23.  Find  the  surface  of  a  sphere  whose  circumference  is  22  ft. 

24.  The  radius  of  a  sphere  is  35  ft.,  and  the  altitude  of  a  zone  is 
15  ft.     Find  the  area  of  the  zone. 

25.  A  baseball  is  9;^  in.  in  circumference.  How  much  leather  is 
required  to  cover  1000  baseballs? 

26.  The  circumference  of  a  dome  in  the  shape  of  a  hemisphere  is 
66  ft.     How  many  square  feet  of  tin  roofing  are  required  to  cover  it  ? 

27.  If  the  ball  on  the  top  of  St.  Paul's  Cathedral  in  London  is  6  ft. 
in  diameter,  what  will  it  cost  to  gild  it  at  7  cents  per  square  inch  ? 

28.  The  altitude  of  the  torrid  zone  on  the  earth  is  about  3200  miles. 
What  is  its  area  in  square  miles,  assuming  the  radius  of  the  earth  to 
be  4000  miles? 

29.  The  edges  of  two  cubes  are  to  each  other  as  2:1,  Compare 
their  surfaces. 

30.  If  each  dimension  of  a  rectangular  parallelopiped  is  multiplied 
by  3,  what  effect  is  produced  on  the  surface  ? 

31.  The  heights  of  two  circular  cylinders  having  equal  bases  are  as 
2:1.     Compare  their  lateral  surfaces. 

32.  The  radii  of  the  bases  of  two  cylinders  having  equal  heights  are 
as  2  :  1.     Compare  their  lateral  surfaces. 

33.  If  the  height  and  the  radius  of  the  base  of  a  cylinder  are  both 
doubled,  what  is  the  effect  on  the  lateral  surface  ? 

34.  Compare  the  lateral  surfaces  of  a  cylinder  and  a  cone  having 
equal  bases  and  equal  altitudes. 

35.  The  lateral  surfaces  of  a  cylinder  and  a  cone  having  equal 
bases  are  equal.     Compare  their  altitudes. 

36.  The  radii  of  two  spheres  are  as  4  : 1.     Compare  the  surfaces. 

37.  The  radius  of  a  sphere  is  7  in.  The  radius  of  the  base  of  a 
cone  is  also  7  in.  What  must  be  the  height  of  the  cone  in  order  that 
its  lateral  surface  may  be  equal  to  the  surface  of  the  sphere  ? 


150  FIRST    STEPS   IN    GEOMETRY 

121.  Volumes  of  the  Common  Solids.  A  unit  of  volume  is  a 
cube  whose  edge  is  equal  to  a  unit  of  length. 

The  volume  of  a  solid  is  the  number  of  units  of  volume  it 
contains. 

The  most  important  units  of  volume  are  : 

The  cubic  inch  (cu.  in.). 

The  cubic  foot  (cu.  ft.)     =1728  cu.  in. 

The  cubic  yard  (cu.  yd.)  =27  cu.  ft. 

The  cubic  centimeter  (ccm). 

The  cubic  decimeter  (cdm),  or  liter  (1)  =  1000  ccm. 

The  cubic  meter  (cbm)  =1000  cdm. 

The  volume  Fof  a  solid  can  be  found  in  the  most  impor- 
tant cases  by  applying  the  following  formulas  : 

Rectangular  Parallelopiped. 

V=  length  X  breadth  X  height. 

Right  Prism  or  Cylinder  of  Revolution. 
V=  base  X  altitude. 

Regular  Pyramid  or  Cone  of  Revolution. 

V=  ^  X  base  X  altitude. 

Any  Frustum  (B  and  b  its  bases). 

V=  i  X  altitude  X  {B  -{- b  +  ^B  x  b). 

Sphere. 

V=  ^ir  X  cube  of  radius. 


EXERCISES   ON   VOLUMES   OF   SOLIDS  151 

EXERCISES 

A  cubic  foot  of  ivater  weighs  62.J^  lb. 

1.  What  is  the  volume  of  a  cube  if  each  edge  is  8  in.  long? 

2.  Find  one  edge  of  a  cube  if  the  volume  is  64  cu.  ft. 

3.  Each  edge  of  a  cubical  tank  measures  5  ft.  6  in.  How  many- 
cubic  feet  of  water  will  it  hold  ?     What  is  the  weight  of  the  water  ? 

4.  How  many  tons  of  coal  will  a  bin  20  ft.  by  16  ft.  by  8  ft.  hold, 
allowing  40  cu.  ft.  to  a  ton  ? 

5.  A  cellar,  the  floor  of  which  measures  12  ft.  by  6  ft.,  is  flooded  to 
the  depth  of  8  in.     Find  the  weight  of  the  water. 

6.  How  many  bricks  9  in.  by  4^  in.  by  3  in.  are  required  to  build 
a  wall  90  ft.  long,  8  ft.  high,  and  18  in.  thick? 

7.  A  book  is  8  in.  long,  6  in.  wide,  and  1^  in.  thick.  A  box  is  3  ft. 
4  in.  long  and  2  ft.  6  in.  wide.  How  deep  must  the  box  be  to  hold 
400  books  ? 

8.  Find  the  weight  of  a  block  of  marble  9  ft.  6  in.  long,  2  ft.  3  in. 
wide,  and  2  ft.  thick,  if  marble  is  2.7  times  as  heavy  as  water. 

9.  Find  the  volume  of  a  regular  triangular  prism  if  the  height  of 
the  prism  is  7  ft.  and  one  side  of  the  base  is  2  ft. 

10.  Find  the  volume  of  a  prism  10  ft.  high  the  bases  of  which  are 
regular  hexagons  each  having  a  side  10  in.  long, 

11.  How  many  cubic  yards  of  earth  must  be  excavated  in  making 
a  cylindrical  well  3  ft.  in  diameter  and  20  ft.  deep?  What  weight  of 
water  will  the  well  contain  when  full  ? 

12.  What  is  the  cost  of  making  a  tunnel  100  yd.  long  whose  section 
is  a  semicircle  with  a  radius  of  10  ft.  at  $6  per  cubic  yard  of  material 
removed  ? 

13.  What  change  in  the  volume  of  a  cylinder  is  produced  by  dou- 
bling its  height?  by  doubling  the  diameter  of  its  base?  by  doubling 
both? 

14.  A  rectangular  sheet  of  tin  is  88  in.  long  and  66  in.  wide.  Find 
the  volume  of  the  cylinder  made  by  rolling  up  the  sheet : 

(1)  so  that  the  height  of  the  cylinder  is  88  in. ; 

(2)  so  that  the  height  of  the  cylinder  is  66  in. 

15.  Find  the  volume  of  a  square  pyramid  if  a  side  of  the  base  is 
40  ft.  and  the  height  is  48  ft. 


152  FIRST    STEPS   IN   GEOMETRY 

16.  Find  the  volume  of  a  pyramid  if  the  height  is  30  ft.  and  the 
base  is  a  regular  hexagon  whose  side  is  6  ft. 

17.  Find  the  volume  of  a  cone  if  the  height  is  28  ft.  and  the  radius 
of  the  base  is  14  ft. 

18.  A  right  triangle  whose  legs  are  6  in.  and  8  in.  long  is  revolved 
about  the  shorter  leg.     Find  the  volume  of  the  cone  thus  generated. 

19.  Find  the  volume  of  the  frustum  of  a  square  pyramid  if  the 
sides  of  the  bases  are  12  in.  and  4  in.  and  the  height  is  6  in. 

20.  The  sides  of  the  bases  of  the  frustum  of  a  square  pyi'amid  are 
16  ft.  and  32  ft.     The  sZawi  height  is  20  ft.     Find  the  volume. 

21.  A  round  stick  of  timber  is  20  ft.  long,  3  ft.  in  diameter  at  one 
end  and  2  ft.  6  in.  at  the  other.  How  many  cubic  feet  of  wood  are 
there  in  the  stick? 

22.  A  bucket  is  16  in.  deep,  18  in.  in  diameter  at  the  top,  and  12  in. 
in  diameter  at  the  bottom.  How  many  gallons  of  water  will  it  hold, 
reckoning  7^  gal.  to  the  cubic  foot? 

23.  Find  the  volume  of  a  sphere  if  the  diameter  is  28  ft. 

24.  Find  the  volume  of  a  sphere  if  the  radius  is  3  ft.  6  in. 

25.  If  one  cubic  inch  of  iron  weighs  a  quarter  of  a  pound,  what  will 
an  iron  ball  1  ft.  in  diameter  weigh  ? 

26.  The  piston  of  a  pump  is  14  in.  in  diameter  and  moves  through 
a  space  of  3  ft.  How  many  tons  of  water  will  be  thrown  out  by 
1000  strokes? 

27.  A  body  is  placed  under  water  in  a  right  cylinder  60  in.  in 
diameter,  and  the  level  of  the  w^ater  is  observed  to  rise  30  in.  Find 
the  volume  of  the  body. 

28.  Compare  the  volumes  of  a  cylinder  and  a  cone  that  have  equal 
bases  and  equal  heights. 

29.  The  edge  of  a  cube  is  10  in.     Find  : 

(1)  the  volume  of  the  largest  cylinder  that  can  be  made  from  it ; 

(2)  the  volume  of  the  largest  oone  ; 

(3)  the  volume  of  the  largest  sphere. 

30.  The  diameter  of  a  wooden  cylinder  and  its  height  are  each 
14  in.     Find: 

(1)  the  volume  of  the  cylinder; 

(2)  the  volume  of  the  largest  sphere  that  can  be  made  from 

the  cylinder. 


ANSWEES 


Page  43.  7.  360.  8.  5400.  9.  36,380'';  5°.  10.  150°;  135°;  120°; 
100°;  60°;  30°.    11.  80°;  70°;  60°;  50°;  30°;  15°.    12.  90°.    13.  45°.   14.  60°. 

Page  54.  3.  c  =  e  :=  £^  =  40° ;  b  =  d=f=h=  140°.  4.  90°.  5.  c  =  e 
=  g  =  60°;  b  =  d=f=h  =  120°. 

Page  55.  5.  4^  mi.  nearly.  6.  6f  mi.  nearly.  7.  144°  37'  30". 
8.  10°  38'.     9.  93°  15'.     10.  33°  54'.     11.  51°  25' 43".     12.  51°  25' 43". 

Page  60.     2.  150°.     3.  80°.     4.  50°. 

Page  61.     6.  45°.     8.  100°^     9.  79°  6'. 

Page  62.     1.  130°.     2.  90°.     3.  90°.     5.  114°. 

Page  63.     2.  40°;  100°.     3.  65°;  65°.     4.  45°. 

Page  64.     6.  60°.     6.  60°;  30°.     7.  40°;  140°;  140°. 

Page  67.     2.  22^.     1.  30°;  60°. 

Page  82.     1.  720°.     2.  1080°.     3.  1440°.     4.  1800°.     5.  9. 

Page  84.     2.   120°;  90°;  72°;  60°;  45°;  36°;  3°  36'.     3.  120°.     4.  135°. 

5.  144°. 

Page  92.     1.110  ft.     2.  88  in.     3.50.2656  cm.     4.42  ft.     5.0.15915  m. 

6.  66,080.  7.  66Sj\  ft.  8.  7  ft.  7yV  in.  9.  1536|f  ft.  10.  2|f  in. 
11.  3ft.  8in.  12.  llin.  13.  7  ft.  7|  in.  15.  45°.  16.  1  mi. ;  57°  16' 22". 
17.  57°  16' 22". 

Page  95.     1.  81  sq.  in.        2.   196  sq.  ft.        3.  4|f  sq.  in.       4.  225  qcm. 
5.  625  qm.     6.  41.2164  qcm.     7.  $68.45.     8.  285,156  sq.  ft.;  6.546  A. 
Page  96.     1.   108  sq.  in.      2.  93  sq.  in.      3.  7|  sq.  in.      4.   12,000  sq.  ft. 

5.  6000  sq.  ft.     6.  192  qcm.     7.  800  qcm.     8.  5  in.     9.   12  in.      10.  2k  in. 

11.  10  ft.     13.  5866f  sq.  yd. 

Page  97.  14.  2312.  15.  280  sq.ft.;  40;  40  yd.  16.  $4233.60.  17.  2  ft. 
8  in.  18.  100  rd.  19.  Doubled ;  doubled  ;  quadrppled.  20.  112,320  sq.  ft. 
21.  Square  by  25  qm.  22.  6  in.  23.  6.32  in.  24.  7650  sq.ft.;  625  sq.  ft. 
25.  400  sq.  ft. 

Page  98.     1.  80  sq.  in.     2.  35  sq.  in.     3.  54.94  qcm. 

Page  99.     1.  55  sq.  in.     2.  l\  sq.  in.     3.  105^i  sq.  ft.     4.  44.72  qcm. 

Page  100.     1.  24  sq.  in.      2.  29^|  sq.  in.      3.  21,780  sq.  ft.      4.  360  ft. 

6.  24  scj.  in. 

Page  101.  1.  54sq.  in.  2.  125  qcm.  3.  589  sq.  ft.  4.  |  sq.  mi. 
5.  560  sq.  ft.     6.  138,564  sq.  ft. 

Page  102.  1.  10.8 sq.  in.  2.  61.92  sq.  in.  3.  1086.3  sq.  ft.  4.  2777.895 
sq.ft.     5.  $207,840.     6.  110.848  sq.  ft.     7.   192  sq.  ft.     8.   1.61ft. 

Page  103.  1.  8  A.  1280  sq.  yd.  2.  1  A.  3104  sq.  yd.  3.  20  A.  4750 
sq.  yd.     4.  10  A.  391  sq.  yd. 

Page  105.  1.  10  cm.  2.  17  cm.  3.  12  cm.  4.  12  cm.  5.  30  ft. 
6.26ft.     7.1200ft.     8.  60mi.     9.  150sq.  in.     10.28.28cm.     11.14.14cm. 

12.  48  sq.ft.  13.  17.32  ft.;  173.2  sq.  ft.  14.  28.28  ft.  15.  14.14  ft, 
16.    30  ft. 

153 


164  ANSWERS 

Page  106.     1.  154sq.  in.     2.  616sq.  ft.     3.  1256.64 qcm.     4.  3.1416 qm. 

5.  1886.5  sq.  ft.        6.   9.625  sq.  ft.        7.   201.0624  qcm.        8.   7857}  sq.  in. 

9.  125.714  A.     10.  5544  sq.  ft.     11.  770  sq.  ft. 

Page  107.  12.  3080  sq.  ft.  13.  |400.  14.  16.  15.  616  sq.  in. 
18.576.72  ft.  19.  3.74  in.  20.6.2  ft.  21.  5.66  in.;  8  in.  22.  74.45  ft. 
23.  7.90  ft.     24.  50.91  A.     25.  207 f  yd.;  204 2  sq.  yd. 

Page  108.  26.  77  sq.  in.  27.  12f  sq.  in.  28.  1283^  sq.  in.  29.  102f 
sq.ft.  30.  462  sq.  in.  31.  77  sq.  in.;  45°.  32.  ^  ;  60°.  33.  4.45  sq.  in. 
34.  896  sq.  ft.     35.  4.5  sq.  ft. 

Page  110.     1.  1:2;  2:1;  2:3;  1:3;  3:7;  7:5. 

Page  111.    1.  1:4;  8:1;  1:6.    3.40.    4.15.    5.4.    6.10.    7.1.5.     8.9. 

Page  112.  2.  320  sq.ft.  3.  33  :  13  ;  33  :  20.  4.  300  sq.ft.  8.  756  sq.ft.; 
24  ft. 

Page  113.  11.10:1.  12.1:2.  13.2:1.  14.2:1.  19.  1:3;  1:9. 
20.  Doubled  ;  quadrupled.  21.  1  :  5  ;  1  :  25.  22.  160  sq.  ft.  23.  ^11.25  ; 
$31.25. 

Page  118.  1.  1:5.  2.  1  : 4.  3.  2  :  3.  4.  21  cm  ;  27  cm.  5.  40  cm; 
50  cm;  1  :  5.     6.  4  : 1.     7.  4:3. 

Page  119.     10.  12.     11.  20. 

Page  123.  1.  7500  sq.  ft.  2.  25  ft.  3.  282.8  ft.  ;  353.5  ft.  1.  1  :  36. 
2.   1:64;   1:8. 

Page  125.     1.  16  in.;  10  in.;  2  in.;  1  in.;  ^  in.;  2  in.;  l^  in.     2.  3|^  mi. 

Page  127.     1.  1287^  ft. 

Page  128.     1.  1838.5  ft. 

Page  130.     1.   142.56  ft.     2.  124  ft. 

Page  131.     1.  75  ft.     2.  70  ft.     8.  205f  f t.     4.  40  m.     5.  160  ft. 

Page  132.     1.  163.92  ft. 

Page  133.     1.  9.987  A. 

Page  134.     1.  15.05  A. 

Page  135.     1.  20. 78  A. 

Page  136.     1.  359  A. 

Page  144.     2.  13  cm.     3.  16  ft. 

Page  148.     1.216  qcm.     2.  lOl^sq.  ft.    3.  558  qcm.    4.  $56.    5.17,774. 

6.  622. 764  sq.ft.  7.  471f  sq.  in.  8.  1005f  sq.  ft.  9.  $415.80.  10.  1800  qcm. 
11.  360  qcm.  12.  622.56  qcm.  13.  264  sq.  f t.  14.  17.08  cm;  321.95 
qcm.  15.  113}  sq.  in.  16.  9605.2  sq.  ft.  17.  320  sq.  in.  18.  2.64  in. 
19.  1037.7  sq.  ft. 

Page  149.  20.  3401.568.  21.  1809.5616  qcm.  22.  154sq.  in.  23.  154 
sq.  ft.  24.  3300  sq.  ft.  25.  189.06  sq.  ft.  26.  693.  27.  $1140.48. 
28.  80,457, 142f.  29.  4:1.  30.  Multiplied  by  9.  31.  2:1.  32.  2:1. 
33.  Multiplied  by  4.     34.2:1.     35.1:2.     36.16:1.     37.  27.11  in. 

Page  151.  1.  512  cu.  in.  2.  4  ft.  3.  166|cu.  ft.;  10,.381.8  lb.  4.  64. 
5.   2995.2  lb.      6.   15,360.      7.   20  in.       8.   7202.52  lb.       9.   12.124  cu.  ft. 

10.  18  cu.  ft.  72  cu.  in.  11.  5^\  cu.  yd.  ;  8825  lb.  12.  $10,476.19. 
14.  301,1861  cu.  in.  ;  401,581^  cu.  in.     15.  25,600  cu.  ft. 

Page  152.     16.  9-35.28  cu.  ft.  17.  5749^  cu.  ft.  18.   402|  cu.  in. 

19.  416cu.  in.  20.  10,949.12  cu.  ft.  21.  119.16  cu.  ft.  22.  12|7. 
23.  ll,498f  cu.  ft.  24.    179|  cu.  ft.  25.    226f  lb.  26.    100.1. 

27.  84,857  }  cu.  in.  28.  ^  :  1.  29.  785|  cu.  in. ;  261if  cu.  in. ;  523^|  cu.  in. ; 
30.  2156  cu.  in. ;  1437^  cu.  in.     The  sphere  is  just  f  as  large  as  the  cylinder. 


II^DEX 


The  numbers  refer  to  the  pages 


Alternate  interior  angles,  54 

Altitude,  57,  58,  59 

Angle,  9 ;  generation  of,  42 ;  right, 
straight,  acute,  obtuse,  42 ;  meas- 
ure of,  44 ;  central,  72 ;  inscribed, 
75 

Angular  units,  42 

Arc,  31 

Area,  93  ;  of  square,  95 ;  of  rectangle, 
96 ;  of  parallelogram,  98 ;  of  tri- 
angle, 99 ;  of  rhombus,  100 ;  of 
trapezoid,  101 ;  of  regular  polygon, 
102  ;  of  any  polygon,  103;  of  circle, 
106. 

Areas  of  similar  figures,  122 

Base,  22,  56,  58,  137,  141,  144 
Bi-symmetric  figures,  48 
Body,  1 

Central  angle,  72 

Chord,  31 

Circle,  31 

Circumference,  31 

Circumscribed  figures,  84 

Compasses,  32  ^ 

Complement  of  an  angle,  43 

Cone,  33  ;  of  revolution,  144 

Cube,  1,  137 

Curved  line,  3  ;  surface,  5 

Cycloid,  41 

Cylinder,  33  ;  of  revolution,  143 


Decagon,  82 

Development,  17,  22,  23,  25 

Diagonal,  58,  82 

Diameter,  31 

Dimension,  1,  57 

Direction,  3 

Distance  between  two  parallels,  53 

Distance  from  point  to  line,  52 

Dividers,  6 

Division  of  circumference,  86 

Dodecagon,  82 

Drawing  exercises,  34 

Drawing  to  scale,  124 

Ellipse,  38 
Equal  figures,  7 
Equivalent  figures,  93 
Exterior-interior  angles,  54 

Face,  2 
Frustums,  145 
Fundamental  problems,  64 

Geometric  body,  26 
Geometric  equality,  7 
Gunter's  chain,  133 

Hexagon,  82 

Horizontal  lines  and  planes,  14 

Icosagon,  82 

Indirect  measurement,  126 


155 


156 


INDEX 


Inscribed  angle,  75 

"        figures,  84 
Intersection,  2 

Length  of  circumference,  91 
Line,  2 
Loci,  37 

Metric  units  of  length,  21 

Model  of  cube,  17 

"      "  rectangular  prism,  23 
"      "  square  prism,  22 
"      "  triangular  prism,  25 

Numerical  measure,  109 

Octagon,  82 

Parabola,  39 

Parallel  lines,  11,  52;  planes,  12 

Parallelogram,  58 

Parallel  perspective,  138 

Path  of  a  moving  line,  30 

"      "         "       point,  29 
Pentagon,  82 
Perimeter,  56 

Perpendicular  lines,  10  ;  planes. 
Pin-ruler  method,  126 
Plane  surface,  5  ;  figure,  6 
Plumb  rule,  16 
Point,  2 

Polygon,  82  ;  regular,  83 
Prisms,  22,  23,  24,  137,  139,  140 
Problems,  64,  69,  89 
Proportion,  111 
Protractor,  45 
Pyramids,  141,  142 

Quadrant,  72 
Quadrilaterals,  58 

Radius,  31 


Kailroad  curves,  79 

Ratio,  109 

Rectangle,  58 

Rectangular  parallelopiped,  137 

"  prism,  23,  137 

Regular  polygon,  83  ;  prism,  137 
Rhomboid,  58 
Rhombus,  58 
Right  angle,  9  ;  prism,  137  ;  cylinder, 

143 
Ruler,  4 ;  ruler  and  triangle,  18 

Secant,  77 

Sector,  72 

Segment,  72 

Similar  polygons,  117 

Sphere,  33,  146 

Spherical  sector,  146  ;  spherical  seg- 
ment, 146 

vSpirit  level,  16 

Square,   10,  58;   square  prism,    22; 
square  pyramid,  142 

Stone  pavements,  88 

Straight  line,  3 

Supplement  of  an  angle,  43 

Surface,  2 
13         Surveying,  133 

Symmetry,  axial,  47  ;  central,  50 

Tangent,  77 

Theorem,  62  ;  of  Pythagoras;  104 

Trapezium,  58 

Trapezoid,  58 

Triangles,  18,  24,  56 

Triangular  prism,  24,  140 

Try  square,  13 

Units  of  length,  19,  21 ;  of  angle,  42 ; 
of  surface,  94  ;  of  volume,  150 

Vertex,  9,  56,  141 

Vertical  lines  and  planes,  14 


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